This is false.  The archetypical family of hyperbolic Coxeter groups are the [hyperbolic triangle groups](https://en.wikipedia.org/wiki/Triangle_group#The_hyperbolic_case)
$$
T(l,m,n) = \langle a,b,c \mid a^2=b^2=c^2=(ab)^l=(bc)^m=(ca)^n=1\rangle
$$
where $l,m,n\geq 2$ and $(1/l)+(1/m)+(1/n)<1$.  Such a group acts as a group of symmetries of a tiling of the hyperbolic plane by congruent triangles with angles of $\pi/l$, $\pi/m$, and $\pi/n$.  Here are some basic facts about these groups:

 *  If we choose a base triangle $T_0$ (a.k.a. the "fundamental chamber"), then each triangle of the tiling can be expressed uniquely as $gT_0$ for some $g\in T(l,m,n)$.

 * As such, the (right) Cayley graph of $T(l,m,n)$ is precisely the dual graph to the triangular tiling, with the identity vertex lying in the triangle $T_0$.

 * The triangular tiling can be obtained by cutting the hyperbolic plane along a countable family of hyperbolic lines.  A path of edges in the Cayley graph is a geodesic if and only if it crosses each line of the tiling at most one time.

 * The Gromov boundary $\partial T(l,m,n)$ is the circle $S^1$, which can naturally be identified with the boundary circle of the hyperbolic plane.  An infinite geodesic path in the Cayley graph represents a point on the boundary if and only if it converges to that point  in the closed unit disk.

Now, let $L$ be the hyperbolic line that separates $T_0$ from $aT_0$, let $z$ be an endpoint of $L$ on the boundary circle, and consider the following two geodesic paths in the Cayley graph:

* The path $\alpha$ that goes from $T_0$ to $aT_0$, and then "follows along" $L$ in the direction of $z$.

* The path $\beta$ that starts at $T_0$ and "follows along" $L$ in the direction of $z$.

To be precise, $\beta$ is the path in the Cayley graph corresponding to the sequence of triangles passed through by the hyperbolic ray from the identity vertex to $z$, and $\alpha$ consists of the dual edge from $T_0$ to $aT_0$ followed by the reflection of $\beta$ across $L$.  Then $[\alpha]=[\beta]=z$, but $\alpha$ starts with $a$ and $\beta$ does not.

**Edit:** By the way, part of the reason I chose the above example was to convey some intuition for hyperbolic Coxeter groups.  However, there are simpler examples available.  For example, consider the group
$$
G = \langle a,b,c \mid a^2=b^2=c^2=(ab)^2=(ac)^2=1\rangle
$$
This is the direct product of $\mathbb{Z}_2$ with the infinite dihedral group $\mathbb{Z}_2*\mathbb{Z}_2$, and its Cayley graph is a bi-infinite "ladder" with $a$ edges as rungs and alternating $b$ and $c$ edges along the sides.  This group is Gromov hyperbolic since it's virtually cyclic (making it an "elementary" hyperbolic group), and the Gromov boundary $\partial G$ is a two-point set corresponding to the two ends of the ladder.

Now, if $\alpha$ is the geodesic path corresponding to the infinite word $a(bc)^\infty = abcbcbc\cdots$ and $\beta$ is the infinite path corresponding to $(bc)^\infty$, then $[\alpha]=[\beta]$ since the two paths go the same direction on the ladder, but $\alpha$ starts with $a$ and $\beta$ does not.