The center of your group $G$ has to be finite (since it is finitely generated, abelian, and has no element of infinite order -- if it did, the group would fail to be word-hyperbolic, since that element together with some other element of $G$ of infinite order would generate a $\mathbb{Z} \oplus \mathbb{Z}$). But now, Lemma 3.3 of http://arxiv.org/pdf/1106.4595 states that no discrete group of motions of an Hadamard manifold ($\mathbb{H}^n$ certainly qualifies) has a finite normal subgroup.

The center of your group $G$ has to be finite (since it is finitely generated, abelian, and has no element of infinite order -- if it did, the group would fail to be word-hyperbolic, since that element together with some other element of $G$ of infinite order would generate a $\mathbb{Z} \oplus \mathbb{Z}$). But now, Lemma 3.3 of http://arxiv.org/pdf/1106.4595 states that no discrete group of motions of an Hadamard manifold ($\mathbb{H}^n$ certainly qualifies) has a finite normal subgroup.

EDIT If the orbifold has cusps, and the center has an element of infinite order $\gamma,$ then all the elements of $G$ fix the fixed point (which has to be on the sphere at infinity) of $\gamma$ (since they all commute with $\gamma$) which contradicts the finite volume assumption. In fact, if $\gamma$ has finite order, it has a fixed point in $\mathbb{H}^n,$ that fixed point has to be fixed by every element of $G,$ and so $G$ is a subgroup of $SO(n)$ (and the quotient is, again, not finite volume) -- the argument before the edit is unnecessary, but is left there for historical reason.