Specifically, I was wondering if the surface was non-compact with infinitely generated free fundamental group, could the surface bundle itself have infinitely generated free fundamental group. In this case, the fundamental group of the bundle is $F_\infty \rtimes_\phi \mathbb{Z}$. On a group theory level, the above can happen; i.e. there exist groups $F_\infty \rtimes_\phi \mathbb{Z} \cong F_\infty$, as was explained to me here:
https://mathoverflow.net/questions/106472/could-f-infty-rtimes-z-be-isomorphic-to-f-infty
However, I'm not sure if this could be carried out with a surface bundle.
Thanks, Kevin
Let $1\to K\to F_\infty\to\mathbb{Z}\to 1$ be a short exact sequence (so that $K$ is free). Let $S$ be a non-compact surface with fundamental group $\pi_1(S)=F_\infty$, and let $X\to S$ be the normal covering space corresponding to $K\leq \pi_1(S)$. The group of deck transformations of $X$ is then identified with $\mathbb{Z}$. We should then be able to build our surface bundle over $S^1$ as $(X\times\mathbb{R})/\mathbb{Z}$, where $\mathbb{Z}$ acts on $X$ by deck transformation and on $\mathbb{R}$ by translation.