Let $X$ be the planar graph
- whose vertices are the points $(i,j)$ for $i \in \mathbb{Z}$ and $j \in \{-1,0,1\}$;
- whose edges connect $(i,0)$ with $(i+1,0)$ for every $i \in \mathbb{Z}$ and $(i,\pm 1)$ with $(i,0)$ for every $i \in \mathbb{Z}$.
In other words, $X$ is a bi-infinite line with top and bottom neighbours added to every vertex.
Of course, $X$ is a locally finite hyperbolic graph (it's a tree). It is not difficult to verify that its isometry group is the unrestricted wreath product $$\mathbb{Z}/2\mathbb{Z}\ \mathrm{wr} \ \mathbb{D}_\infty = \left( \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z} \right) \rtimes \mathbb{Z}/2\mathbb{Z}.$$ Each $\mathbb{Z}/2\mathbb{Z}$ from the product corresponds to an isometry that swaps the top and bottom neighbours $(i,1)$ and $(i,-1)$ for some $i \in \mathbb{Z}$ but fixes all the other vertices. The $\mathbb{Z}$ corresponds to the obvious translation. And the right $\mathbb{Z}/2\mathbb{Z}$ corresponds to a left-right reflection.
The boundary $\partial X$ of $X$ contains only two points, and the action of $\mathrm{Isom}(X)$ on $\partial X$ corresponds to the projection onto the right factor $\mathbb{Z}/2\mathbb{Z}$. Consequently, the kernel of the action is the unrestricted wreath product $$\mathbb{Z}/2\mathbb{Z} \ \mathrm{wr} \ \mathbb{Z}:= \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z},$$ which is uncountably infinite.
The key point is that hyperbolicity has no control on small scale geometry. By using the same trick of adding top and bottom neighbours, you can start from your favourite hyperbolic graph $Y$ and create a new hyperbolic graph $Y^+$ (quasi-isometric to $Y$) whose isometry group contains an uncountably infinite subgroup $\prod \mathbb{Z}/2\mathbb{Z}$ in the kernel of its action on $\partial Y^+$.