The question is a little too vague to have a really satisfactory answer, but obviously descent
is omnipresent in (modular) representation theory. 

For example, the basic result that two representations of a group $G$ on some $k$-vector spaces are isomorphic iff they become isomorphic on some extension $L$ of $k$ is an instance of faithfully flat descent.

Or to take a less trivial example, let $R$ be a d.v.r with residual field $k$ algebraically closed (or just separably closed, or finite, or...) and of field of fraction $K$. Let $\rho$ be a representation of a group $G$ on some extension $L$ of $K$, whose characteristic polynomials of element of $g$ are in $R[x]$. (In characteristic zero, you can replace
that condition by "character takes values in $R$").
A natural question in this context is if you can "descend" $\rho$ to a representation over $R$. For example, the answer is yes when there exists a representation of $G$  over $k$
which is absolutely irreducible and whose characteristic polynomial agree with the reduction of those of $\rho$. The proof uses a lot of faithfully flat 
descent via the theory of Azumaya algebras.

Edit: to answer Martin's comment, I precise how the faithful flatness plays a role in the first example. 

Lemma: Let $G$ be a group. Let $k$ be a commutative noetherian ring, $k'$ a flat $k$-algebra. Let $M$, $N$, be  $k[G]$-module, finitely generated as $k$-modules. let $M'=M \otimes_k k'$ and $N'=N\otimes_k k'$.
Then $Hom_{k[G]}(M,N) \otimes_k k' = Hom_{k'[G]}(M',N')$. Moreover if $k'$ is a faithfully flat $k$-algebra, $Hom_{k[G]}(M,N)$ is non-zero if and only if $Hom_{k'[G]}(M',N')$ is non-zero.

Pf: Let $g_1,\dots,g_n$ be a finite family of elements of $g$ generating the image of $k[G]$ in $End_{k}(N)$.
Let $t_i$ be the linear endomorphism of $H:=Hom_{k}(M,N)$ given by $\phi \mapsto \phi g_i - g_i \phi$. Consider the map $H \rightarrow H^n$ given by the $t_i$. The kernel of this map
is $Hom_{k[G]}(M,N)$. the kernel of the same map tensorized by $k'$ is $Hom_{k'[G]}(M',N')$.
Hence the first result by flatness. The second is clear by faithful flatness.

So for example, if $k$, $k'$ are field, $M$, $N$ simple $k[G]$-modules, one retrieve
the classical assertion that $M \simeq N$ off $M' \simeq N'$. But I hope I have shown that 
the essence of this assertion is faithfully flat descent.