I think the answer to your question is negative (i.e. $(F_p^\times \ltimes F_p,F_p)$ does not have relative property (T)), by using the combinatorial proof that semidirect products of amenable groups are again amenable (see Proposition 5 of my notes at http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/ ).  A bit more specifically, let $S_p$ be a bounded set of affine transformations on $F_p$, and let $D_p$ be the associated set of dilations on $F_p$, which is then a bounded subset of $F_p^\times$.  By the abelian nature of the dilation group, we can then construct a nontrivial set $F$ of the dilation group $F_p^\times$ which is $99.9\%$-invariant with respect to the dilations of $D_p$ (thus $|dF \Delta F| \leq 10^{-3} |F|$ for all $d \in D_p$), basically by constructing a medium-sized generalised geometric progression using the dilations in $D_p$ as generators.  Note that one can make the set $F$ of bounded size (i.e. independent of $p$.)  Let $T$ be the set of all translations $t$ such that $SF$ intersects $Ft$ (i.e. the translations in $F^{-1} S_p F$).  This is a set whose size is controlled by the size of $S_p$ and of $F$, and in particular is still bounded uniformly in $p$.  We can then construct a moderately large (but still of size uniformly bounded in $p$) set $E$ of translations which is $99.9\%$-invariant with respect to any of the translations of $T$.  The set $U := EF$ will then (for $p$ large enough) be a non-trivial $99\%$-invariant set with respect to $S_p$, by the argument given in my notes above.