This is already false for $G={\rm GL}(1,F)$. In that case a cuspidal irreducible representation is a smooth character $\chi$ of $F$. The contragredient is $\chi^{-1}$. We have $\chi \sim \chi^{-1}$ iff $\chi^2$ is unramified. There are easy counter-examples.

Let us give another counter-example in higher rank. Take $G={\rm PGL}(n,F)$ and $\pi$ compactly induced from an irreducible smooth representation $\lambda$ of ${\rm PGL}(n,O_F )$, $O_F$ the ring of integers of $F$. Assume that $\lambda$ is lifted from a cuspidal representation $\sigma$ of ${\rm PGL}(n,k_F )$, where $k_F$ is the residue field of $F$ (so $\pi$ is of level $0$).
Then $\lambda$ is a type for the Bernstein block of $\pi$ in the sense of Bushnell and Kutzko's type theory. If $\pi\sim {\tilde \pi}$ then $\lambda$ and $\tilde \lambda$ are types for the same Bernstein block. By a standard result of type theory for ${\rm PGL}(n,F)$, $\lambda$ and $\tilde \lambda$, whence $\sigma$ anb $\tilde \sigma$, are isomorphic. By Green's parametrization, $\sigma$ is attached to a regular character $\chi$ of $l^\times$, trivial on $k_F^\times$, where $l/k_F$ is a degree $n$ extension. Then $\sigma\sim {\tilde \sigma}$ is equivalent to $\chi$ and $\chi^{-1}$ belonging to the same ${\rm Gal}(l/k_F )$-orbit. It is then an easy exercise to build up a counter-example.

hasno (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$). $\endgroup$ – LSpice Mar 23 '18 at 13:49