$\newcommand{\HH}{\mathbb H}$ $\newcommand{\RR}{\mathbb R}$ $\newcommand{\ZZ}{\mathbb Z}$ $\newcommand{\SL}{\mathop {\rm SL}}$
In your question $X = G \backslash \HH$ is given as a quotient. Also, $G$ is a finitely generated free group and a subgroup of $\SL(2,\RR)$. As $G$ is convex co-compact it follows that $G$ is discrete. In this case $\HH$ is the universal cover of $X$, the group $G$ is the so-called "deck-group", and there is a non-canonical isomorphism of $G$ with $\pi_1(X)$. (Choosing basepoints doesn't make the isomorphism canonical.)
The last paragraph of your question doesn't make much sense. The usual action of $\SL(2,\ZZ)$ on $\HH$ is not convex co-compact. So it is not a good example to think about. Here is a better example which fits your situation very tightly:
Suppose that $G$ is the free group of rank one -- that is, $G \cong \ZZ$. Suppose that $G$ is generated by a hyperbolic isometry $\gamma$. Let $A_\gamma \subset \HH$ be the axis of $\gamma$ acting on $\HH$. So $A_\gamma$ is a copy of $\RR$, topologically, and $G$ acts on $A_\gamma$ as $\ZZ$ acts on $\RR$, by translation. So the quotient $g = G \backslash A_\gamma$ is a circle. Also $X = G \backslash \HH$ is a hyperbolic annulus with two "flaring ends". To make this precise, note that $g \subset X$ is an essential loop in the annulus $X$. Note that $X - g$ has two components $L$ and $R$ (coming from the left and right sides of $A_\gamma$ in $\HH$). $L$ is again an annulus with one boundary on $g$. Also, $L$ has an exponentially flaring metric (as you move away from $g$) so the other boundary of $L$ is "at infinity".
Returning to $\HH$, note that the limit set of $G$, $\Lambda_G$, is exactly two points: the endpoints of $A_\gamma$. Let $\Omega_G = \partial_\infty \HH - \Lambda_G$ be the domain of discontinuity of $G$. Then $\Omega_G$ is two open sub-arcs of $\partial_\infty \HH$. As before $G$ acts on each of these by translation. The quotients are the circles at infinity for $L$ and $R$ respectively. These two circles at infinity are the Gromov boundary of the hyperbolic annulus $X$.
Finally, in this example the convex core of $X = G \backslash \HH$ is exactly the circle $g = G \backslash A_\gamma$ that we started with.
This discussion generalizes. If $G \subset \SL(2,\RR)$ is any (torsion free) group acting convex co-compactly on $\HH$ then the fundamental group of $X = G \backslash \HH$ is isomorphic to $G$. Also, the Gromov boundary of $X$, $\partial_\infty X$, is homeomorphic to $G \backslash \Omega(G)$. Each component of $\partial_\infty X$ is a circle - there is one for each flaring annular end of $X$. All ends are of this type - convex co-compactness rules out cusps.