Are the complements of two homeomorphic compact, connected subsets of $\mathbb{R}^2$ homeomorphic?

Suppose that $A,B$ are subsets of $\Bbb{R}^2$. If $A$ and $B$ are homeomorphic and $A$ and $B$ are compact and connected, are their complements homeomorphic?

Yes. Let's work in the sphere $S^2$ for convenience. Let $A$ be a connected compact proper subset of $S^2$. Its complement is an open (noncompact without boundary) surface $M'$; its complement in $\mathbb R^2$ is $M' - \{\infty\}$. This is a disjoint union of connected noncompact surfaces without boundary; each of these must have free fundamental group.

For manifolds $M$, the Cech cohomology $\check H^*(M)$ is isomorphic to the singular cohomology $H^*(M)$. By Alexander duality, and the fact that $\check H_0(A) = 0$ (everything is happening in the reduced world!), we see that $H^1(M') = 0$. (The $0$th Cech homology more precisely counts quasicomponents, but for compact Hausdorff spaces, quasicomponents are the same as components.) Because for a disjoint union $H^1(X \sqcup Y) \cong H^1(X) \times H^1(Y)$, we see that every component of $M'$ is simply connected. By the uniformization theorem, each simply connected component is homeomorphic to $\mathbb R^2$. Now delete the point at infinity; this makes one of your copies of $\mathbb R^2$ into an annulus $S^1 \times (-1,1)$.

So your $M$ is homeomorphic to $S^1 \times (-1,1) \sqcup \kappa \mathbb R^2$, where by $\kappa \mathbb R^2$ I mean the disjoint union of $\kappa$-many copies of $\mathbb R^2$. What's left is to determine $\kappa$. But again we just need to use Alexander duality, because $H^0(M') \cong \prod_{\kappa+1} \mathbb Z$, and this is isomorphic to $\prod_{\kappa'+1} \mathbb Z$ iff there's a bijection $\kappa \to \kappa'$.

So $M$ is determined by something even coarser than the homeomorphism type of $A$: all one needs is the 0th and 1st Cech homologies $\check H_*(A)$.