Riffing off of Ben Steinberg's answer, which amounts to the statement that a tree is not quasi-isometric to the hyperbolic plane, here's a proof which doesn't require knowing anything about ends.
Tile $H^2$ by regular $2g$-gons which are fundamental domains for the action of $\pi_1 X$. The boundaries of these $2g$-gons form the a Cayley graph $\Gamma$ of $\pi_1 X$. If $\pi_1 X$ were a free group then $\Gamma$ would be quasi-isometric to a tree, and so there would exist constants $C \ge 0$ and $s \in (0,1)$ such that every closed edge path $\gamma$ in $\Gamma$ can be written as a concatenation $\gamma = \gamma_1 * \gamma_2$ so that $Length(\gamma_1), Length(\gamma_2) \ge s Length(\gamma)$, and the initial and terminal endpoints of $\gamma_1$ have distance $\le C$ in $\Gamma$. But for sufficiently large $r > 0$, choosing $\Gamma$ to be a closed edge path that stays within a uniform distance of the radius $r$ circle in $H^2$, we get a contradiction.
