Is there a residually finite non-elementary hyperbolic group whose profinite completion is boundedly generated? Is there a residually finite hyperbolic group $G$ that is not virtually cyclic, such that there exists finitely many procyclic closed subgroups $C_1, \dots, C_n$ of the profinite completion $\hat{G}$ of $G$ satisfying the property that for every $g \in \hat{G}$ there exist $c_1 \in C_1, \dots, c_n \in C_n$ with $g = c_1 \cdots c_n$ ?
It is known that $G$ itself can't be boundedly generated.
 A: I'm fairly certain that no example is known. Of course, it's a famous open problem whether every hyperbolic group is residually finite.  This turns out to be equivalent to many other questions about the profinite topology on hyperbolic groups (see, for instance, https://arxiv.org/abs/0802.0709 and http://www.math.uiuc.edu/~kapovich/PAPERS/dani.dvi.gz ).  It's very tempting to conjecture that the existence of a non-elementary hyperbolic group with boundedly generated profinite completion would imply the existence of a non-residually finite hyperbolic group.
Most of the hyperbolic groups that we know to be residually finite are known to be so because they are cubulable (i.e. the fundamental group of a compact, non-positively curved cube complex). Agol's theorem, which resolved the virtual Haken conjecture, implies that these groups are residually finite.  In this case, we can say more.   Any non-elementary, cubulable, hyperbolic group always has a finite-index subgroup that retracts to a non-abelian free group. It follows that the profinite completion virtually surjects a non-abelian profinite free group, and hence can't be boundedly generated.
In the other direction, one would like candidate positive examples; here, the first example to look at is a cocompact lattice $\Gamma$ in $Sp(n,1)$. The congruence subgroup problem is open for such lattices. This asks whether the profinite completion of $\Gamma$ is equal to the congruence completion, meaning the inverse limit of all the congruence quotients of $\Gamma$. If so, then $\Gamma$ is said to satisfy the congruence subgroup property (CSP).
The lattice $\Gamma$ is linear, and hence certainly residually finite.  Nevertheless, Lubotzky observed that, if the CSP holds for $\Gamma$, then there is a non-residually finite hyperbolic group.  And, indeed, the OP points out in comments that the congruence completion of $\Gamma$ is known to be boundedly generated.
In summary, we can say that bounded generated of the profinite completion is closely tied to the existence of a non-residually finite hyperbolic group. On the one hand, in cubulable groups, which are very "robustly" residually finite, the profinite completion is not boundedly generated. And in quaternionic lattices, the most plausible source of a non-residually finite hyperbolic group would also lead to a non-elementary, residually finite hyperbolic group with boundedly generated profinite completion.
