I think this can be seen directly. Let's work over $G=PGL(n)$, so I don't have to keep repeating "modulo the center". Recall that supercuspidal representations can be realized as subrepresentations in $L^2(G)$ consisting of compactly supported functions.
So we can define an intertwining integral from $\pi$ into the Whittaker space by
$$\Big(g\rightarrow f(g)\Big)\longrightarrow \Big(g\rightarrow\int_N \bar\psi(n)f(ng)\ dn\Big)$$
("compactly supported" gives convergence).
Let $W$ denote the Whittaker function of an $f\in\pi$. If the support of $f$ is $C\subset G$, then the support of $W$ is $NC$. To see this, take $g\notin NC$. So, for each $n\in N$, $ng\notin C$. So $f(ng)=0$ for all $n$. Hence $W(g)=0$. Since $C$ is compact, $NC$ is compactly supported modulo $N$.
[Added: I realized I was tacitly assuming that for some $f$, the above integral is not identically zero. Since $f$ has compact support, the integral being identically zero implies that $f$ lies in the kernel of the twisted Jacquet functor. Since the representation has a Whittaker module, the twisted Jacquet module must be non-zero, hence there is an $f$ whose integral does not vanish.]
I don't know a reference off-hand for this basic fact (it also follows from the fact that the functions in the Kirillov model of a supercuspidal representation have compact support, which is in a lot of sources), though Bushnell and Henniart's "Supercuspidal Representations of $GL_n$: Explicit Whittaker Functions" gives a more, well, explicit version of it. [Added: I just realized that this is the final result of Casselman-Shalika's Whittaker function paper.]