Let $X$ be a topological space, $Y$ be a set and let $q: X \to Y$ be a quotient map. In other words $q$ is surjective and $Y$ is equipped with the final topology induced by $q$. Let $A \subseteq X$ be a subspace of $X$ equipped with subspace (relative) topology and let $p = q|_A$, i.e.: $$p: A \to Y \ ,x \mapsto q(x) $$
Now forget about $X$ and define the quotient space $Y'$ respect to the map $p$.
When $Y$ and $Y'$ are homeomorphic?
First of all we need $Y = Y'$ as sets. We can show that $p$ is surjective, i.e. it covers all its range $Y$. Thus we have to check that $p^{-1}(q(x))$ is not empty for all $x \in X$. In other words $q^{-1}(q(x)) \cap A$ has not to be empty.
Is this sufficient?
Or we have to require topological properties on $A$?
PS: I drastically changed the text of the question. I used quotient maps (as used in topology) instead or equivalence relation in order to simplify the conditions.
PSS: I found the answer in the proposition quoted in this question: A question about the restriction of quotient maps to subsets of domain.
Your condition is not sufficient.
For example, let $X=\mathbb R$ with the usual topology and $x\sim y$ iff $x-y\in\mathbb Z$.
Then $X/{\sim}$ is a circle $S^1$.
However, let $A=[0,1)$. This satisfies your condition of containing at least one representative of each equivalence class, but the restriction of $\sim$ to $A$ is the identity and therefore $A/{\sim}$ is $[0,1)$ itself, which is distinctly different from the circle.