Short answer:
This condition is weaker than continuity, and as stated it is highly dependent on a (hidden) choice: It is similar to checking continuity "only along a certain path". If one demands it to be true for all such choices, one recovers continuity. It might be interesting to investigate: For how many choices does one have to impose this condition to ensure "full" continuity?
Long answer:
For each $k=0,1,2, \dots$ let $C_k$ be a set of coefficients, i.e. $p$ elements in $\mathbb Z_p$ whose residues mod $p$ make up $\mathbb Z_p/(p)$.
For a fixed sequence of coefficient sets $\mathbf C:= (C_k)_k$, each $x\in \mathbb Z_p$ can be written as $\sum_0^\infty c_k(x) p^k$, where each $c_k(x) \in C_k$ is uniquely determined by $x$.
The most “standard” choice is $C_k =\{0,1, \dots p-1\}$ for all $k$, let's call this $\mathbf C_{st}$. Another one commonly used are the Teichmüller representatives, and again "the same" for all $k$. But the uniqueness statement above still holds if we fix different coefficients for each $k$.
For each fixed $\mathbf C = (C_k)_k$, one has the truncation maps $tr_\mathbf C^n : \mathbb Z_p \rightarrow \mathbb Z_p$, $x \mapsto \sum_{k=0}^n c_k(x) p^k$.
Let $D$ be any metric space, and let's call a map $f:\mathbb Z_p \rightarrow D$ weakly $\mathbf C$-continuous at $x$ if
$$f(x) = \lim_{n\to \infty} f(tr_{\mathbf{C}}^n(x))$$
Your condition is weak $\mathbf C_{st}$-continuity.
Observation 1: If $f$ is continuous at $x$, then it is weakly $\mathbf{C}$-continuous at $x$ for all $\mathbf{C}$.
Example 1: A function which is weakly $\mathbf C_{st}$-continuous everywhere, but continuous only at $x\neq 0$: $$f(x)= \begin{cases} 0 \text{ if } x \neq 0 \\1 \text{ if } x=0 \end{cases}$$
(where $1$ and $0$ can be replaced by any two different values in our codomain $D$ which, to avoid trivialities, contains more than one element). More generally, if you choose finitely many natural numbers $a_1, ..., a_k \in \mathbb N$, you can define an $f$ to have arbitrary values at those, while being (another) constant on $\mathbb Z_p \setminus \{a_1, ..., a_k\}$, and such $f$ is still weakly $\mathbf C_{st}$-continuous everywhere, but continuous only on $\mathbb Z_p \setminus \{a_1, ..., a_k\}$.
Example 2: A function which is weakly $\mathbf C_{st}$-continuous everywhere, but continuous only on $\mathbb Z_p \setminus \mathbb N$. (You gave this example in a comment, with codomain $\mathbb Z_q$, but as Qiaochu Yuan has noted, analogous constructions work e.g. for $D= \mathbb R$. This is a "smoothened" upgrade of example 1.)
$$f(x) = \begin{cases} 0 \text{ if } x \notin \mathbb N \\ q^{d_{st}(x)} \text{ else,} \end{cases} $$
where $d_{st}(x)$ is the number of nonzero $c_k(x)$ in the $\mathbf C_{st}$-expansion of $x$.
It seems like the natural numbers stand out here. They should not, because we do not do any arithmetic. They do because they are special with respect to the choice of coefficients $\mathbf C_{st}$: They are the ones which are "eventually stable under truncation". For other choices, one finds analogous constructions with analogous sets:
Terminology: For given $\mathbf C$, let $Fin(\mathbf{C}) \subset \mathbb Z_p$ be the set of elements $x\in \mathbb Z_p$ such that there is $n$ such that $x = tr_{\mathbf{C}}^m(x)$ for all $m\ge n$. For $\mathbf C$ where all $C_k$ contain $0$, these are exactly those $x$ that can be written as finite sums $\sum_{k=0}^m c_k(x) p^k$. For example $Fin(\mathbf C_{st}) = \mathbb N$.
Observation 2: For given $\mathbf C$, every (!) $f$ is weakly $\mathbf C$-continuous at all $x \in Fin(\mathbf C)$.
Observation 3: The function in example 1 is weakly $\mathbf{C}$-continuous for every $\mathbf{C} = (C_k)_k$ such that $0$ is contained in all $C_k$. (A condition which is often imposed on coefficient systems. Note the function is not $(C_k)_k$-continuous if e.g. all $C_k$ contain $p$ and $-1$.)
Example 3: A $\mathbf C$ (with all $C_k$ "the same") such that the function in example 2 is not weakly $\mathbf{C}$-continuous e.g. at $x=1$.
Let $p=2$ and $\mathbf C = (C_k)_k$ with $C_k = \{0,3\}$ for all $k$. We have $c_k(1) = \begin{cases} 3 \text{ if } k= 0,1,3,5,7, ... \\ 0 \text{ if } k=2,4,6,8,... \end{cases}$
In particular, $tr_\mathbf{C}^{2n-1}(1) = tr_\mathbf{C}^{2n}(1) = 1+2^{2n+1}$ for all $n \ge 1$.
But that means $d_{st}(tr_{\mathbf{C}}(1)) = 2$ for $n \ge 1$, while obviously $d_{st}(1)=1$, so $q^{d_{st}}$ is not $\mathbf{C}$-continuous at $x=1$.
Upshot: The function cooked up from $d_{st}$ in example 2 is weakly $\mathbf C_{st}$-continuous because it relies on the same "standard" coefficient choice. Analogous functions $d_{\mathbf C}$, which count nonzero coefficients in $\mathbf C$-expansions, would give rise to examples analogous to example 2 which are weakly $\mathbf C$-continuous everywhere albeit not continuous on $\mathbb Z_p \setminus Fin(\mathbf C)$. But as example 3 shows, these have no reason to be weakly $\mathbf C'$-continuous for a different choice $\mathbf C'$.
I think of this as checking continuity only along "a very special path": When choosing $\mathbf C$, you choose to check if $f$ is continuous at $x$ when $x$ gets approximated by elements of $Fin(\mathbf C)$, but not "from other directions".
But if one considers all possible choices, one recovers full continuity:
Observation 4: Let $f:\mathbb Z_p \rightarrow D$ and $x \in \mathbb Z_p$ be given. If $f$ is weakly $\mathbf{C}$-continuous at $x$ for all $\mathbf C$, then $f$ is continuous at $x$.
Proof: Let $x_0, x_1, x_2, ...$ be an arbitrary but fixed sequence converging to $x$. We need to show $\lim_{n\to \infty} f(x_n) =f(x)$. The idea is to construct a $\mathbf C = (C_k)_k$ such that the given sequence $(x_n)_n$ (or a big enough part of it) is contained in $Fin(\mathbf C)$. More precisely, I claim we can find a subsequence $x_{n_k}$ and coefficient sets $C_k$ such that $tr_{\mathbf C}^k(x) = x_{n_k}$ for all $k$. In fact, we choose both recursively via the following properties:
Now as wanted, $tr_{\mathbf C}^k(x) = x_{n_k}$ for all $k$, so that $f$ being weakly $\mathbf{C}$-continuous at $x$ means that $\lim_{k\to \infty} f(x_{n_k}) = f(x)$.
This argument only relies on $x_n \to x$, so it can be applied to any subsequence of $x_n$ as well. Hence the sequence $(f(x_n))_n$ (in the metric codomain $D$) has the property that each of its subsequences has another subsequence that converges to $f(x)$. This implies (cf. 1) that $f(x_n)$ itself converges to $f(x)$, q.e.d.
Open question: For $f$ to be continuous at $x$, is it sufficient that $f$ is weakly $\mathbf{C}$-continuous at $x$ for every "constant" $\mathbf{C}$, i.e. where all coefficients come from the same set $C_0=C_1=C_2=\dots$? -- The construction in the proof of observation 4 generally will give different $C_k$ for each $k$. But so far I see no example of an $f$ that has a discontinuity at an $x$ which cannot be "detected" by some constant $\mathbf C$ (at that specific $x$).
