This isn't true. Here's one easy construction. For some $g \geq 2$, let $\Sigma_g$ be a closed oriented genus $g$ surface and let $f\colon \Sigma_g \rightarrow \Sigma_g$ be a pseudo-Anosov homeomorphism that acts trivially on $H_1(\Sigma_g)$ and fixed a point $\ast \in \Sigma_g$. There are many constructions of such $f$. By Thurston's hyperbolization theorem, the mapping torus $M_f$ is hyperbolic. Now let $S$ be the universal abelian cover of $\Sigma_g$. Using the fixed point of $f$, you can lift $f$ to a homeomorphism $\tilde{f}\colon S \rightarrow S$. The mapping torus $M_{\tilde{f}}$ is then a cover of $M_f$, so it is also hyperbolic. The associated short exact sequence is $$1 \longrightarrow \pi_1(S) \longrightarrow \pi_1(M_{\tilde{f}}) \longrightarrow \mathbb{Z} \longrightarrow 1,$$ and $\pi_1(S)$ is an infinite rank free group.

This isn't true. Here's one easy construction. For some $g \geq 2$, let $\Sigma_g$ be a closed oriented genus $g$ surface and let $f\colon \Sigma_g \rightarrow \Sigma_g$ be a homeomorphism representing a pseudo-Anosov mapping class that acts trivially on $H_1(\Sigma_g)$. There are many constructions of such pseudo-Anosov mapping classes (for example, this paper shows that suitable "random" mapping classes acting trivially on homology are pseudo-Anosov). For simplicity, isotope $f$ such that it fixes a basepoint.

By Thurston's hyperbolization theorem, the mapping torus $M_f$ is hyperbolic. Now let $S$ be the universal abelian cover of $\Sigma_g$. Using the fixed basepoint of $f$, you can lift $f$ to a homeomorphism $\tilde{f}\colon S \rightarrow S$. The mapping torus $M_{\tilde{f}}$ is then a cover of $M_f$, so it is also hyperbolic. The associated short exact sequence is $$1 \longrightarrow \pi_1(S) \longrightarrow \pi_1(M_{\tilde{f}}) \longrightarrow \mathbb{Z} \longrightarrow 1,$$ and $\pi_1(S)$ is an infinite rank free group.