My understanding of the updated version of the question, about (Gromov-)hyperbolic groups, is that, at least in the torsion-free case, it would indeed follow from the Cannon conjecture, as the OP suggests. The case with torsion would also follow if every hyperbolic group is residually finite, and indeed it may follow unconditionally.

Let's recall:

Cannon conjecture: If $\Gamma$ is a hyperbolic group and the Gromov boundary of $\Gamma$ is homeomorphic to $S^2$ then $\Gamma$ is a Kleinian group (i.e. acts properly by isometries on hyperbolic 3-space).

Suppose, now, that $\Gamma$ is hyperbolic, torsion-free, and acts properly and cocompactly by homeomorphisms on $\mathbb{R}^3$. Since the action is proper and $\Gamma$ is torsion-free it follows that the action is free, and so the quotient $M=\Gamma\backslash\mathbb{R}^3$ is a closed, aspherical (in particular, irreducible) 3-manifold with (Gromov-)hyperbolic fundamental group. Therefore, by Theorem 4.1 of Bestvina and Mess, the Gromov boundary of $\Gamma$ is homeomorphic to $S^2$, and so the Cannon conjecture implies that $\Gamma$ is Kleinian as required.

It's well known that a residually finite hyperbolic group is virtually torsion-free. Therefore, if one is willing to assume that $\Gamma$ has a torsion-free (wlog normal) subgroup $\Gamma_0$ of finite index then, as above, $\Gamma_0$ is the fundamental group of a hyperbolic 3-manifold $M$. By a theorem of Gabai, the action of the deck group $\Gamma/\Gamma_0$ on $M$ is by isometries, and it follows that $\Gamma$ itself is also Kleinian. (Note that if $\Gamma$ does not have a torsion-free subgroup of finite index then it is a non-residually finite hyperbolic group, and so resolves a different famous open problem!)

It may be possible to eliminate the assumption that there is torsion-free subgroup of finite index. I want to say that a properly discontinuous, cocompact, topological action of $\Gamma$ on $\mathbb{R}^3$ is enough to imply that $\Gamma$ is a rational $PD_3$ group, and then apply Theorem 4.8 from this paper of Bestvina (which requires a hypothesis on the orientation character, and I don't know if this is satisfied) to deduce that the boundary of $\Gamma$ is a rational homology 2-sphere. I think this is enough to deduce that the boundary is in fact homeomorphic to the 2-sphere, from which one could then conclude again using the Cannon conjecture. But there are several potential holes in this line of reasoning and I don't have time to chase up all the references and try to fill them in myself.

However, perhaps it's worth mentioning that Brian Bowditch -- who is a colleague of the OP -- should certainly be able to say whether or not the line of reasoning in the final paragraph is valid.

My earlier attempt at an answer was a bit of a mess -- let me have another go.

The hypotheses of the question introduce several technical difficulties, but I'm unsure which are crucial and which can be relaxed. Certainly, if we're willing to relax them slightly then we can get a positive answer, so I'll give an answer under certain hypotheses that seem reasonable to me.

The usual discreteness hypothesis in this context is not an absence of accumulation points, but proper discontinuity, and it seems to me that this is the natural way to generalise Pardon's theorem. With the hypothesis of absence of accumulation points (which @YCor rightly points out is strictly weaker) it's not clear to me what happens even for smooth actions. Apologies if this strictly weaker properness hypothesis is the point of the question (but I don't see a connection with Cannon's conjecture).

So let's suppose that $\Gamma$ is a hyperbolic group acting properly discontinuously and cocompactly by homeomorphisms on $\mathbb{R}^3$. To keep things simple, let's also assume that $\Gamma$ has a (wlog normal) torsion-free subgroup $\Gamma_0$ of finite index. 

Since the action is properly discontinuous and $\Gamma_0$ is torsion-free, the action of $\Gamma_0$ is free and so the quotient $M_0=\Gamma_0\backslash\mathbb{R}^3$ is a closed topological 3-manifold.

By Moise's theorem $M_0$ has a smooth structure, and now $M_0$ is an aspherical 3-manifold whose fundamental group has no $\mathbb{Z}^2$ subgroups, so $M_0$ admits a hyperbolic metric by the geometrisation theorem. This metric pulls back to realise the action of $\Gamma_0$ as an action by isometries on $\mathbb{H}^3$.

Pardon's theorem shows that the action of the finite deck group $\Gamma\backslash\Gamma_0$ on $M_0$ can be approximated by smooth actions, and a theorem of Gabai implies that this action is isotopic to an action by isometries. As a result, the action of the whole group $\Gamma$ on $\mathbb{H}^3$ is also by isometries, as desired.