When does convergence in quotient space $X / {\sim}$ induce convergence in $X$
Such a $q$ is called "sequence covering" and is defined in this paper by Siwiec from 1971.
Proposition 2.4 from that paper might be useful: an almost open map defined on a first countable space is sequence covering.
Here $f: X \rightarrow Y$ is almost open, iff for all $y \in Y$ there is some $x$ with $f(x) = y$ such that, for every open $O$ in $X$ with $x \in O$, $y \in \operatorname{int}(f[O])$. Any open map satisfies it.
A simple example (2.6) shows things can go wrong easily: Let $X$ be the set $\{\frac{1}{n}: n \in \mathbb{N}^+\} \cup \{0\} \cup \{2+\frac{1}{n}: n \in \mathbb{N}^+\} \cup \{2\}$ as a subset of the reals, and let $\sim$ be the equivalence relation that only identifies $0$ and $2$ and leaves the rest alone.
Then both $X$ and its quotient are metric compact, and $q$ is perfect (closed map, all inverse images of points are finite (at most size 2) hence compact). But $q$ is not sequence covering, as the sequence $[1],[2+\frac{1}{2}],[\frac{1}{3}],[2+\frac{1}{4}],[\frac{1}{5}],\ldots$ converges to $[0] = [2]$ but there is only one choice of lifting the sequence and this does not converge. So even only one non-trivial class between extremely nice spaces can fail this property.
Googling "sequence covering maps" will give you some more papers with generalisations of 2.4. The paper I linked to contains some more counterexamples where we cannot drop some assumption.
You can't expect this to happen even in some very nice cases. For instance, suppose $X=[0,1]$ and $\sim$ is the equivalence relations whose equivalence classes are $\{0,1\}$ and $\{x\}$ for each $x\in(0,1)$. Then in $X/\sim$, the sequence $\{1/2\},\{1/3\},\{2/3\},\{1/4\},\{3/4\},\{1/5\},\{4/5\},\dots$ converges to $\{0,1\}$. However, there is only one possible choice of a representative of each equivalence class, and the sequence $1/2,1/3,2/3,1/4,3/4,\dots$ you get does not converge in $X$.
Here is the most general condition I can find in which it does work. Suppose $X$ is first-countable and $\sim$ is an equivalence relation such that the closure of any saturated set is saturated (for instance, this is true if the equivalence classes of $\sim$ are the orbits of a group of homeomorphisms of $X$). Fix any point $x\in p$, and let $U_1\supseteq U_2\supseteq U_3\supseteq\dots$ be a descending local base at $x$. For each $N$, let $S_N=\{n:p_n\cap U_N=\emptyset\}$. Suppose $S_N$ is infinite for some $N$. Let $A=\overline{\bigcup_{n\in S_N}p_n}$. Then $A$ is the closure of a saturated set and hence saturated, so its image in $X/{\sim}$ is also closed. Since $S_N$ is infinite and $p_n\to p$, this means the image of $A$ contains $p$, and so $p\subseteq A$. But this is impossible, since $U_N$ is a neighborhood of $x$ that witnesses that $x\not\in A$.
Thus $S_N$ is finite for all $N$. We can now choose a sequence of points $x_n\in p_n$ such that $x_n\in U_N$ whenever $N\leq n$ and $n\not\in S_N$. By the finiteness of $S_N$, this sequence is eventually in each $U_N$, and hence converges to $x$.
To connect this argument to Henno Brandsma's answer, note that the closure of any saturated set is saturated iff the quotient map is an open map. Indeed, the closure of any saturated set is saturated iff the interior of any saturated set is saturated. If this holds and $U\subseteq X$ is open, then the interior of the saturation of $U$ is saturated and contains $U$ and hence is the entire saturation of $U$. That is, the saturation of $U$ is open, so its image in the quotient is open. Conversely, if the quotient map is open and $S\subseteq X$ is saturated, the saturation of the interior of $S$ is open and hence equal to the interior of $S$ since $S$ is saturated. That is, the interior of $S$ is saturated.