Is there a contractible hyperbolic 3-orbifold of finite volume? Let $\mathbb{H}^3:=\operatorname{SO}(3,1)/\operatorname{O(3)}$.
Is there a lattice $\Gamma$ in $\operatorname{SO}(3,1)$ such that
\begin{equation}
X:=\mathbb{H}^3/\Gamma
\end{equation}
is contractible? Of course, we need to assume that $\Gamma$ has torsion.
 A: As pointed out by Moishe Kohan in the comments below, the following doesn't answer the question as asked, because my group $\Gamma$ is not contained in $SO(3,1)$.  Anyway, here is an easy description of a $\Gamma$ that contains orientation-reversing elements that has all the other properties that are asked for.
By the intermediate value theorem, there is a regular dodecahedron in hyperbolic 3-space whose dihedral angles are all $90^\circ$.  (The angles for a Euclidean regular dodecahedron are greater than $90^\circ$ and the angles for an ideal regular dodecahedron are $60^\circ$.)  The group generated by the reflections in the sides of this dodecahedron is a discrete group $\Gamma$ with the property that $\mathbb{H}/\Gamma$ is contractible, since it's just a copy of the dodecahedron.
A: Yes. For example, let $M$ be the figure-eight knot complement. So $M$ is a hyperbolic manifold with volume a bit more than 2.  The manifold $M$ has a two-fold symmetry $\tau$ that fixes, pointwise, a pair of geodesics $\alpha$ and $\beta$.  The symmetry $\tau$ extends over the figure-eight knot to give an isometry of $S^3$.  In $S^3$ the arcs $\alpha$ and $\beta$ (together with two points of the knot) form a great circle - $\tau$ is a 180 degree rotation about this great circle.
We take the boundary of a small regular neighbourhood of the figure eight to get a "cusp torus". This torus meets the great circle in four points.  The quotient of the cusp torus is a euclidean orbifold of type $S^2(2, 2, 2, 2)$.  Using the above, we find that $M/\tau$ is homeomorphic to an open ball; the images of $\alpha$ and $\beta$ are the orbifold locus in $M/\tau$ and "look like" a tangle in the open ball.  Here is a picture of what I get this way:


EDIT: Every orientable hyperbolic orbifold, with contractible underlying space, is essentially of this "type".  That is, it is a labelled graph embedded in the open three-ball.
To see this: tameness and contractibility implies that the boundary is a single two-sphere and the underlying manifold is an open three-ball.  Since $\Gamma$ is a lattice it has finite covolume.  Thus the number of cone points in the boundary orbifold is either three or four.  Their markings make the boundary into a euclidean orbifold.  The orbifold locus in the interior of the three-ball may have vertices - the boundary of a regular neighbourhood of any of these is a spherical orbifold.
In the example above, there are no internal vertices, and the boundary orbifold is $S^2(2, 2, 2, 2)$.

I vaguely remember a paper of Hatcher's that lists the first few (arithmetic?) hyperbolic orbifolds. [EDIT: Here is a link to that paper (now that Hatcher has kindly reminded me of the title).]
