The answer is always. If $G$ is an anisotropic group over a non-archimedean local field $k$, then $G(k)$ is compact. The elements of $G(k)$ are semi-simple (if $char k =0$) and hence their conjugacy classes are closed in the Zariski topology and therefore also in the $k$-topology. 

[Edit] To round it off, the question may easily be reduced to the case when $G$ is semi-simple and simply connected; hence $G$ is a product of simply connected $k$-simple groups $G_i$, and by a restriction of scalars argument, $G$ may be assumed to be absolutely almost simple simply connected. In this case, the classification (see Tits' article in the AMS symposia series on algebraic groups, discrete subgroups...) says that $G$ is $SL_1(D)$ for a central division algebra over $k$ or $SU_1(D)$ where $D$ is a central division algebra with an involution of the second kind. The semi-siple conjugacy classes may now be described explicitly.