Let us consider a surface $S$ and a subset $T\subset S$, where $T\cong D^2$ are homeomorphic and $D^2$ is an open disc. Let $T\cong Q\subset S$ be homeomorphic.
Are $S\setminus T $ and $S\setminus Q$ homeomorphic?
If not, under which circumstances is it possible?
I think, if $T\subseteq Q$ and there exists an open $\epsilon$-neighbourhood around $Q$ which we call $U_\epsilon(Q) $, s.t. $U_\epsilon(Q)\cong D^2$, then $S\setminus Q\cong S\setminus T$.
Because: $U_\epsilon(Q)\setminus Q$ is homeomorphic to an annulus. But also $Q\setminus T$ should be homeomorphic to an annulus. But then the implication is clear.
In an analog way we see: If there exist an $\epsilon$-neighbourhood and $\delta $-neighbourhood around $T$ and $Q$ which we call $U_\epsilon(T)$ and $U_\delta (Q)$, s.t. they are homeomorphic to $D^2$, then $S\setminus Q\cong S\setminus T$.
But I can't imagine a way solving it without such neighbourhoods for a general surface. Maybe there is a way solving the problem for compact surfaces. But this is more intuition than real knowledge.