I believe the answer is NO in general. For $n>3$ one obtains a counter-example by taking any maximal torus $T$ corresponding to a partition $\{k, n-k\}$ of $n$ such that $2\leq k \leq n-2$. Clearly $T$ is a $C_{pp}$ group because it doesn't contain any $p$-elements. On the other hand it clearly doesn't preserve an $(n-1)$-dimensional space. Finally it is clear that its order is not divisible by a primitive prime divisor of $q^n-1$ or $q^{n-1}-1$.

Similarly if $n=3$ and $3$ divides $q-1$, then the normalizer of a split torus provides a counter-example. If $3$ does not divide $q-1$, then I'd have to think some more.

If $n=2$, then a counter-example is given by a Sylow $p$-subgroup where $p$ is the characteristic of the field. 

**Remark**: I assume, by the way, that your definition of $C_{pp}$-group should exclude central elements of $SL_n(q)$. Because, clearly, if the center of $SL_n(q)$ is non-trivial then any $C_{pp}$-subgroup of $SL_n(q)$ must contain no $p$-elements. The counter-examples given above for $n\geq 3$ still work even with the definition you've given, but, still, I imagine that the correct definition of a $C_{pp}$-group should be that the centralizer of any $p$-element has form $PZ$ where $P$ is a $p$-group and $Z=Z(SL_n(q))$.