It appears to me that $\bar{X}^H$ is disconnected in general, will always contain the connected contractible piece $X^H$, and have other components corresponding to intersections of $H$ with rational parabolics of $G$. For example, any $\mathbb{Q}$-irreducible finite subgroup $F$ of $G_{\mathbb{Z}}$ will only have trivial intersection with a rational parabolic, and hence $\bar{X}^F=X^F$ is contractible (by the usual CAT(0) arguments).

It appears to me that $\bar{X}^H$ is disconnected in general, will always contain the connected contractible piece $X^H$, and have other components corresponding to intersections of $H$ with rational parabolics of $G$. For example, any $\mathbb{Q}$-irreducible finite subgroup $F$ of $G_{\mathbb{Z}}$ will only have trivial intersection with a rational parabolic, and hence $\bar{X}^F=X^F$ is contractible (by the usual CAT(0) arguments). In case $F$ meets some rational parabolics $\{P_\nu\}_\nu$, then $F< G_{\mathbb{Z}}$ is a $\mathbb{Q}$-reducible representation, and stabilizes some finite set $\{ \nu\}$ of vertices in the rational Tits building. It does not seem possible to describe in general the possible adjacency relations of these $F$-invariant vertices, i.e. $\bar{X}^F \setminus X^F$ may or may not be a simplex (e.g. for integral subgroups of rational Weyl group representations) and may perhaps be a pair of adjacent edges o--o--o.