$\DeclareMathOperator\tr{tr}\DeclareMathOperator\Int{Int}\DeclareMathOperator\Cent{Cent}$The OP (@ClarkLyons) gave a lovely generalisation of an answer of @diracdeltafunk over on MSE. I believe that the technique can be pushed still further to handle maps to any group $H$, as in the original question. The argument has no new ideas from me.
Suppose that $\phi : G \to H$ is any homomorphism. For $g \in G$, write $G\cdot g$ for the $G$-conjugacy class through $g$; and similarly $H\cdot h$ for $h \in H$. Write $f : H \to \mathbb Z_{\ge 0}$ for the function given by $$ f(h) = (\#H\cdot h)^{-1}\sum_{g \in \phi^{-1}(H\cdot h)} (\#G\cdot g)^{-1} $$ for all $h \in H$.
The Fourier transform $\hat f : \rho \mapsto (\#H)^{-1}\sum_{h \in H} f(h)\tr \rho(h)$ satisfies $$ (\#H)\hat f(\rho) = \sum_{h \in H} (\#H\cdot h)^{-1}\sum_{g \in \phi^{-1}(H\cdot h)} (\#G\cdot g)^{-1}\tr \rho(h) = \sum_{g \in G} (\#G\cdot g)^{-1}\tr (\rho \circ \phi)(g), $$ which is the inner product of the character of $\rho \circ \phi$ with the character $g \mapsto \#\Cent_G(g)$ of the permutation representation of $G$ acting on itself by conjugation, so that $\hat f(\rho)$ is non-negative, for all irreducible complex representations $\rho$ of $H$.
By column orthogonality of the character table, we have that $$ f(h) = \sum_{\rho \in \hat H} \hat f(\rho)\overline{\tr \rho(h)} $$ for all $h \in H$, so that $$ f(h) = \lvert f(h)\rvert \le \sum_{\rho \in \hat H} \lvert\hat f(\rho)\rvert\cdot\lvert\tr \rho(h)\rvert \le \sum_{\rho \in \hat H} \hat f(\rho)\tr \rho(1) = f(1) $$ for all $h \in H$.
In words, upon multiplying both sides of the inequality $f(h) \le f(1)$ by $\#H\cdot h$, we obtain that the number of $G$-conjugacy classes (in $G$) that map into the $H$-conjugacy class of $h$ is bounded by the product of the size of the $H$-conjugacy class of $h$ and the number of $G$-conjugacy classes in the kernel of $\phi$.