Here is a counterexample for the second question. In fact, I'll show that when $A = M_2(\mathbb{C})$, there is a positive non-linear $\phi: A \to \mathbb{C}$ s.t. $\phi$ is linear on all maximal abelian subalgebras of $A$ (without assuming the subalgebra in question is self-adjoint). This is clearly enough to yield a counterexample.
We first need to consider what is the structure of a maximal abelian subalgebra $B$ of $A$. Clearly, $\mathbb{C} \subsetneq B$. Thus, we may pick a non-scalar $T \in B$. By considering the Jordan normal form of $T$, we have, after changing bases, either,
$$T = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}$$
for some $\alpha \neq \beta$. Let $S = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $[T, S] = 0$ iff,
$$\begin{pmatrix} \alpha a & \alpha b \\ \beta c & \beta d \end{pmatrix} = TS = ST = \begin{pmatrix} \alpha a & \beta b \\ \alpha c & \beta d \end{pmatrix}$$
As $\alpha \neq \beta$, this happens iff $b = c = 0$. That is, the commutant of $T$ consists precisely of diagonal matrices, which form an abelian algebra. As $B$ is maximal abelian, $B$ must be the algebra of diagonal matrices. To put it another way, $B = \mathbb{C}p \oplus \mathbb{C}(1 - p)$ for some not necessarily self-adjoint rank-one idempotent $p$.
Or,
$$T = \begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix}$$
Let $S = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $[T, S] = 0$ iff $[e_{12}, S] = [T - \alpha, S] = 0$ iff,
$$de_{12} + ce_{11} = e_{12}S = Se_{12} = ae_{12} + ce_{22}$$
This happens iff $a = d$ and $c = 0$, iff $S = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$. It is straightforward to check all such $S$ form an abelian algebra. Again, as $B$ is maximal abelian, $B$ must consist of all such $S$. To put it another way, $B = \mathbb{C} \oplus \mathbb{C}x$ for some nonzero nilpotent $x$ (in this case, simply $x \neq 0$ and $x^2 = 0$).
Now, using Zorn's lemma, we may choose a maximal set $P$ of (not necessarily self-adjoint) rank-one idempotents s.t. whenever $p \in P$, we have $1 - p \notin P$. By maximality of $P$, we easily see that for each rank-one idempotent $p$, we have either $p \in P$ or $1 - p \in P$, and exactly one of the above holds. We now define $\phi$ by,
$$\phi(ap + b(1 - p)) = a$$
when $a, b \in \mathbb{C}$ and $p \in P$, and,
$$\phi(a + bx) = a$$
when $a, b \in \mathbb{C}$ and $x$ is nilpotent. By Jordan normal form decomposition, we see that these two cases cover all possible elements of $A$. We observe that $\phi$ is well-defined. Indeed, if,
$$ap + b(1 - p) = c + dx$$
for some $p \in P$ and nonzero nilpotent $x$. As the LHS is a matrix whose eigenspaces span the entire $\mathbb{C}^2$, the RHS must satisfy the same condition, so by uniqueness of Jordan decomposition, we must have $d = 0$. But then $ap + b(1 - p) = b + (a - b)p = c$, so $a = b = c$ and both definitions of $\phi$ yield the same result.
If,
$$a + bx = c + dy$$
for some nonzero nilpotents $x, y$. Then the LHS has unique eigenvalue $a$, while the RHS has unique eigenvalue $c$. Thus, $a = c$ and the definition of $\phi$ yields a unique result.
Finally, if,
$$ap + b(1 - p) = cq + d(1 - q)$$
for some distinct $p, q \in P$. Then by definition of $P$ we must have $q \neq p, 1 - p$. Again, by uniqueness of Jordan decomposition, we must have $a = b = c = d$, so again the definition of $\phi$ yields a unique result.
Now, we check that $\phi$ is linear on all maximal abelian subalgebras. This is nearly just the definition: Let $B \subset A$ be maximally abelian. Then as we have seen, either $B = \mathbb{C}p \oplus \mathbb{C}(1 - p)$ for some $p \in P$, in which case,
$$\phi(ap + b(1 - p)) = a$$
is clearly linear on $B$, or $B = \mathbb{C} \oplus \mathbb{C}x$ for some nonzero nilpotent $x$, in which case,
$$\phi(a + bx) = a$$
is also clearly linear on $B$. But $\phi$ is not linear on the entirety of $A$. Indeed, both $e_{12}$ and $e_{21}$ are nonzero nilpotents, so $\phi(e_{12}) = \phi(e_{21}) = 0$, but $e_{12} + e_{21}$ has distinct eigenvalues $1$ and $-1$, so $e_{12} + e_{21} = p - (1 - p)$ for some rank-one idempotent $p$. Either $p \in P$ or $1 - p \in P$, so,
$$\phi(e_{12} + e_{21}) = \phi(p - (1 - p)) = 1 \text{ or } -1$$
Either way, $\phi(e_{12} + e_{21}) \neq 0 = \phi(e_{12}) + \phi(e_{21})$, so $\phi$ is not linear.
Finally, we check that $\phi$ is positive. For any $T \in A$ positive, we have $T = \alpha p + \beta (1 - p)$ for some rank-one idempotent $p$ and $\alpha, \beta \geq 0$. Again, either $p \in P$ or $1 - p \in P$, so $\phi(T)$ is either $\alpha$ or $\beta$. In both cases, we have $\phi(T) \geq 0$.
