When are quotients of homeomorphic spaces homeomorphic?

Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $\sim$ and $\approx$ and $f:X\to Y$ is a homeomorphism such that $x\sim y$ if and only if $f(x)\approx f(y)$, then $f$ passes to the quotient as $X/_\sim \cong Y/_\approx$.

Proof goes roughly like this: define $\widetilde{f}([x]_\sim)\doteq [f(x)]_\approx$. We have that $\widetilde{f}$ is well-defined and injective because of the condition relating $\sim$ and $\approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $\widetilde{f}$ and $\widetilde{f}^{-1}$ follow from the universal property of the quotient topologies applied to the relations $\widetilde{f}\circ\pi_{\sim}=\pi_{\approx}\circ f$ and $\widetilde{f}^{-1}\circ \pi_{\approx}=\pi_\sim\circ f^{-1}$ with the continuity of $f$ and $f^{-1}$.

I think you are correct. Here is a sketch of the argument:

$\phi:=\pi_Y\circ f$ is a continuous surjection: $X\to Y/\approx$ and it induces a bijection: $X^*=\left \{ \phi^{-1}([y]): [y]\in Y/\approx \right \}\overset{\overline\phi}{\rightarrow} Y/\approx ,\ $ which is a homeomorphism if and only if $\phi$ is a quotient map.

As $\phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/\sim$ and $Y/\approx$ will be homoeomorphic if and only if $X^*$ and $X/\sim$ are.

This means that the map $[x]\mapsto f^{-1}\circ \pi^{-1}_Y([y])$ must be a homeomorphism. That is, $f([x])\mapsto \pi^{-1}_Y([y])$ must be a homeomorphism.

But the above map is well-defined if and only if $x\sim x'\Rightarrow f(x)\approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.

For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $R\rightrightarrows X$, where $R\subset X\times X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $R\rightrightarrows X$ and $S\rightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.

This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:X\to Y$ induces a natural transformation between the diagrams $X\times X\rightrightarrows X$ and $Y\times Y\rightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $R\times R\to S\times S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.