$\newcommand{\RR}{\mathbb R}$ $\newcommand{\HH}{\mathbb H}$ $\newcommand{\Ends}{\mathop{\rm Ends}}$
Let $F$ be an infinite genus surface, properly embedded in $\RR^3$. Then up to homeomorphism of surfaces (not ambient homeomorphisms) there are uncountably many such surfaces.
Proof. Note that $\RR^3$ is homeomorphic to $\HH^3$. So we may consider surfaces properly embedded in $\HH^3$ where there is more "room". Let $\HH^2$ be a totally geodesic plane in $\HH^3$. Let $T = T_3$ be a proper embedding of the regular $3$--valent tree into $\HH^2$. Take a small neighborhood of $T$ in $\HH^3$ to get a strange embedding of a three-ball $B$ into $\HH^3$.
Now, at every vertex of $T$ drill out a small tube from $B$ and take the boundary of the resulting infinite genus handlebody to get an infinite genus surface $F = F(T)$, properly embedded in $\HH^3$. Note that the topological space $\Ends(F)$, the ends of $F$, are an invariant of the homeomorphism type of $F$. In this example, $\Ends(F)$ is homeomorphic to the Cantor set.
By taking a subtree $T' \subset T$ we can obtain another surface $F' = F(T')$. As before, the $\Ends(F')$ is homeomorphic to the Gromov boundary of $T'$. Finally, since there are uncountably many pairwise non-homeomorphic compact subsets of the Cantor set, we find the required uncountable set of pairwise non-homeomorphic embedded surfaces. QED
This answer was partly prompted by Agol's answer to a previous question.
