Not necessarily... In the example you give - a ball in three dimensions with Euclidean metric - there are continuous families of periodic orbits of all numbers of reflections $n\geq 2$ (polygons and stars on planes passing through the origin). But a typical perturbation of the metric would break the symmetry and leave only a finite number of periodic orbits for each $n$. So most of the original periodic orbits are unstable in this sense.
Edit: In response to the question edit: Persistence of a periodic orbit is a local question, depending only on the metric and boundary properties on/near the orbit. It cannot "see" the global symmetry if any. So, the unperturbed manifold need not be symmetric away from the periodic orbit. Furthermore, an isolated non-hyperbolic periodic orbit may also vanish under perturbation of Hamiltonian systems, in for example a saddle-center bifurcation. Such isolated non-hyperbolic orbits may be obtained easily in convex billiards by tuning the local boundary curvatures. The generic situation should be that each orbit is stable to perturbations of the metric, but that any given perturbation destroys infinitely many long periodic orbits.