I think so. The key property used in the proof is $$Res^G_H Ind^G_H V=\bigoplus_{s\in H\setminus G/H} Ind^H_{H_s}V_s,$$ where $V_s$ is the twist of $V$ by $s$, and $H_s=sHs^{-1}\cap H$, and its proof is more or less manifestly characteristic-free. Other than that, instead of characters and scalar products $\langle \chi_U,\chi_W\rangle$ you can use the actual spaces of intertwiners $Hom_G(U,W)$, and finally instead of saying that irreducibility of $W$ means the intertwining number $\dim Hom_G(W,W)$ is equal to 1 (which is true over an algebraically closed field of characteristic zero), you can say that irreducibility of $W$ means that $Hom_G(W,W)$ is a division algebra (which is true in the semisimple case, so under the assumption you want to make).
Let me, following a comment of Darij Grinberg, make the last step a bit more clear. There is a natural homomorphism of algebras $Hom_H(V,V)\to Hom_G(Ind^G_HV, Ind^G_HV)$. The induced representation $Ind^G_HV$ is irreducible if and only if this map is an isomorphism of algebras. However, to conclude that it is sufficient to show that it is an isomorphism of vector spaces, which we do without trouble via the Frobenius reciprocity.
I think so. The key property used in the proof is $$Res^G_H Ind^G_H V=\bigoplus_{s\in H\setminus G/H} Ind^H_{H_s}V_s,$$ where $V_s$ is the twist of $V$ by $s$, and $H_s=sHs^{-1}\cap H$, and its proof is more or less manifestly characteristic-free. Other than that, instead of characters and scalar products $\langle \chi_U,\chi_W\rangle$ you can use the actual spaces of intertwiners $Hom_G(U,W)$, and finally instead of saying that irreducibility of $W$ means the intertwining number $\dim Hom_G(W,W)$ is equal to 1 (which is true over an algebraically closed field of characteristic zero), you can say that irreducibility of $W$ means that $Hom_G(W,W)$ is a division algebra (which is true in the semisimple case, so under the assumption you want to make).