For a unital algebra $A$, and a two-sided ideal $I$ of $A \otimes A$, does there exists a universal quotient algebra $B$ of $A$, with map $q: A \twoheadrightarrow B$, such that $q \otimes q: A \otimes A \twoheadrightarrow B \otimes B$ factors through $(A \otimes A)/I$?
i.e. if there exists another quotient algebra $C$ of $A$ such the map $f: A \otimes A \twoheadrightarrow C \otimes C$ factors through $(A \otimes A)/I$, then the map $A \to C$ factors through $q$?
The answer is "not always". Below, I will give two explicit counterexamples and two explicit examples, and afterwards (this will be the main part) give a general criterion for the existence of this universal quotient. I assume that we work with algebras over a field $K$. This ensures that the tensor product over $K$ preserves monomorphisms.
a) When $I = 0$, the quotient exists and is just $A$.
b) When $A = K[X]$ and $I = \langle X \otimes X \rangle $ the universal quotient does exist: It is the universal $K$-algebra generated by an element $a$ such that $a \otimes a = 0$. This is equivalent to $a = 0$. So the answer is $K$ with $a=0$.
a) Let $A = K[X,Y]$ and $I = \langle X \otimes Y \rangle \subseteq A \otimes A$. Here the universal quotient would be a universal $K$-algebra $B$ generated by two commuting elements $x,y \in B$ with $x \otimes y = 0$ in $B \otimes B$. We must have $x \neq 0$, witnessed by the (non-universal) example $1 \otimes 0 = 0$ in $K \otimes K$. Similarly, we must have $y \neq 0$. But linear algebra tells us that $x,y \neq 0$ implies $x \otimes y \neq 0$, a contradiction.
b) Let $A = K[X]$ and $I = \langle X \otimes 1 - 1 \otimes X \rangle$. The universal quotient would be a universal $K$-algebra $B$ generated by an element $x$ satisfying $x \otimes 1 = 1 \otimes x$. By considering $1 \otimes 1 = 1 \otimes 1$ in $K \otimes K$, we see $x \neq 0$ and $1 \neq 0$ in $B$. Therefore, $x \otimes 1 = 1 \otimes x$ means $x = \lambda 1$ for some $\lambda \in K^*$, but $\lambda$ cannot be chosen universally (unless $K = \mathbb{F}_2$ of course, in which case $K$ with $1 \in K$ is the universal quotient).
Assume that a universal quotient $q : A \twoheadrightarrow B$ exists for an ideal $I \subseteq A \otimes A$. The first step is to show that the universal property, initially stated only for surjective homomorphisms $A \to C$, actually holds for all homomorphisms of algebras.
Let $f : A \to C$ be any algebra homomorphism such that $f \otimes f : A \otimes A \to C \otimes C$ vanishes on $I$, we need to show that $f$ factors uniquely through $q$. Uniqueness is clear since $q$ is assumed to be surjective. Existence is clear by assumption when $f$ is surjective, but if $f$ is not surjective, consider the image decomposition of $f$ as $f' : A \twoheadrightarrow C'$ followed by $i : C' \hookrightarrow C$. Since $i \otimes i$ is injective, then $f' \otimes f'$ vanishes on $I$. Hence, $f'$ factors through $q$. And this shows that also $f$ factors through $q$.
As always, this universal property can also be stated in terms of a representable functor. Here, we consider the functor $F_I : \mathbf{Alg}_k \to \mathbf{Set}$, a subfunctor of $\hom(A,-)$ actually, that maps an algebra $C$ to the set of homomorphisms $f : A \to C$ such that $f \otimes f$ vanishes on $I$. By what we saw above, $F_I$ is representable if $q$ exists. The converse is also true, since it can be checked that every representing object $A \to B$ of $F_I$ must be surjective.
Every representable functor preserves all limits, and by Freyd's representability criterion, the converse is also true in our case, since the (regular) quotients of $A$ provide a solution set for $F_I$. So we should analyse the condition that $F_I$ preserves all limits, or equivalently, equalizers and products (see here for the equivalence). But by a similar argument as above it is easy to check that $F_I$ preserves equalizers anyway.
Hence, a universal quotient exists for $I$ if and only if $F_I$ preserves products. This means that for a family of homomorphisms $(f_i : A \to C_i)_{i \in I}$, if each $f_i \otimes f_i : A \otimes A \to C_i \otimes C_i$ vanishes on $I$, then $f \otimes f : A \otimes A \to \prod_{i \in I} C_i \otimes \prod_{i \in I} C_i$ vanishes on $I$, where $f : A \to \prod_{i \in I} C_i$ is the homomorphism induced by $f_i$.
The natural map $\prod_{i \in I} C_i \otimes \prod_{i \in I} C_i \to \prod_{i,j \in I} C_i \otimes C_j$ is injective (since we are working over a field). Therefore, $f \otimes f$ vanishes on $I$ if and only if each $f_i \otimes f_j : A \otimes A \to C_i \otimes C_j$ vanishes on $I$.
So the condition reduces to: If two homomorphisms $f : A \to C$, $g : A \to C'$ are given such that $f \otimes f$ and $g \otimes g$ vanish on $I$, then also $f \otimes g$ has to vanish on $I$ (and by symmetry then also $g \otimes f$). Now, the kernel of $f \otimes g$ is equal to $\ker(f) \otimes A + A \otimes \ker(g)$, see here. Therefore:
General Criterion. The universal quotient exists for $I \subseteq A \otimes A$ if and only if for all ideals $J,J' \subseteq A$ with $I \subseteq J \otimes A + A \otimes J$ and $I \subseteq J' \otimes A + A \otimes J'$, we also have $I \subseteq J \otimes A + A \otimes J'$.
I am not sure if this condition can be simplified further.