Let me try to muddle my way through. In positive characteristic the Gelfand trick will still tell you that the Hecke algebra of $H$-biinvariant functions $A=Func(H\backslash G/H,k)$ is commutative.
Then, by general nonsense, the Hecke algebra is isomorphic to $End_G(P)$, where $P=k(G/H)=Ind_H^G (k_{triv})$ is the permutation representation.
Since $P$ is not semisimple, its structure can still be complicated. The sort of information you need to conclude that it is a Gelfand pair is that $Soc(P)$ or $P/Rad(P)$ is multiplicity-free. There is no way you can conclude that about $Soc(P)$.
But, here, Doc, is a happy ending for your opera: it follows from Nakayama Lemma that $$End(P/Rad(P))\cong A/Rad(A),$$ forcing the semisimple module $P/Rad(P)$ to be multiplicity free.
Let $V$ be an irreducible $G$-module. By Frobenius-Nakayama-Fudd Reciprocity, $$ Hom_H(k_{triv},V)=Hom_G(P,V)=Hom_G(P/Rad(P),V) $$ must be at most one-dimensional! And so is $$ Hom_H(V,k_{triv})=Hom_H(k_{triv},V^\ast). $$