I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$
Suppose that $M$ is a non-compact, connected, oriented three-manifold without boundary, with universal cover homeomorphic to $\RR^3$. Suppose that $F$ is a compact, connected, oriented surface with genus at least one. Suppose that the fundamental groups of $M$ and $F$ are isomorphic. Is $M$ homeomorphic to $F \times \RR$?
The answer is "no". We can build a counterexample using the paper Some examples of exotic non-compact three-manifolds by Scott and Tucker. Here I closely follow their wording and notation, starting at the bottom of page 488.
Consider their example manifold $M_5$. This is a three-manifold with boundary a closed surface $F$. The universal cover of $M_5$ is homeomorphic to $\RR^2 \times \RR_{\geq 0}$, and the inclusion of $F$ into $M_5$ is a homotopy equivalence. However, $M_5$ is not "almost compact" (that is, it is not tame), and $M_5$ is not homeomorphic to $F \times \RR_{\geq 0}$.
Now double $M_5$ across its boundary to obtain $M = D(M_5)$. This does not change the fundamental group. Note that $M$ has two ends, separated by (the image of) $F$. By Scott-Tucker, neither end is tame. Thus $M$ is not homeomorphic to $F \times \RR$. However, the universal cover of the double is (in this case) the double of the universal cover; thus it is $\RR^2 \times \RR$, as desired.