When $R$ is the Eilenberg-MacLane spectrum of a Noetherian ring, thick subcategories are in bijection with specialization-closed subsets of $\mathrm{Spec}\ \pi_0(R)$.  Such a thick subcategory is generated by a single compact object iff the specialization-closed subset is actually Zariski-closed (and in that case a generator is given by $H(\pi_0(R)/I)$, for $I$ any ideal corresponding to the closed set).   To say that this holds for all thick subcategories imposes a rather strong condition on the ring $\pi_0(R)$; I believe it's actually equivalent to $\mathrm{Spec}\ \pi_0(R)$ having only finitely many points.  (Note that although this condition does hold for $p$-local spectra and the corresponding "Spec" has infinitely many points, one for each height, this situation is badly non-Noetherian.)

The same story holds more generally if the graded ring $\pi_*(R)$ is Noetherian and stratifies the category of $R$-modules, in the sense of [Benson-Iyengar-Krause][1].  For instance, this automatically holds if $\pi_*(R)$ is a regular ring concentrated in even degrees.

Here's a sketch of a proof that if $R$ is a Noetherian ring such that any specialization-closed subset of $\mathrm{Spec}\ R$ is closed, then $\mathrm{Spec}\ R$ is finite.  A specialization-closed set is just a union of closed sets, so this implies that any set consisting only of closed points is closed.  But then quasicompactness implies there can only be finitely many closed points.  We can now look at prime ideals of dimension (coheight) 1, and by a similar compactness argument there can be only finitely many of them.  Continuing by induction on dimension, we get that for each dimension, $R$ can only have finitely many primes of that dimension.  But $\mathrm{Spec}\ R$ has to be finite-dimensional (since, say, there are only finitely many maximal ideal and the localization at any maximal ideal is finite-dimensional), so it can only have finitely many points in total.


  [1]: http://arxiv.org/abs/0910.0642