Suppose $S$ is a non-compact Riemann surface in $\mathbb R^3$ that has no boundary and has genus zero (i.e. its fundamental group is generated only by its ends at infinity). A typical example would be a boundary of a tubular neigborhood of a tree, with all leaves on the sphere at infinity.

Is it true that one of the 3-manifolds into which $S$ divides $\mathbb R^3$ is homeomorphic to $\mathbb R^3$? (The answer is clear for the example given above, but how is it in general?)