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A remark on the (lack of) $p$-adic valued Haar measure: there can exist topologically irreducible continuous representations of pro-$p$ groups (e.g. $GL_n(\mathbb Z_p)$ on infinite-dimensional $p$-adic Banach spaces. This is pretty strong evidence that there is no simple analogue of Haar measure in the $p$-adic situation. (Remember that the existence of Haar measure means that this is impossible for representations of compact groups on real or complex spaces: topologically irreducible representations are necessarily finite-dimensional.)

Another remark, on semi-simplicity: It is conjectured that the representations of the absolute Galois group $G_K$ of $K$ occuring on the $p$-adic etale cohomology of a smooth proper variety over a number field $K$ is semi-simple. (This is a part of the so-called Tate conjecture.) As far as I know, this is very wide open other than in the abelian variety case (and hence in the case of $H^1$ in general, since $H^1$ can always be thought of as the $H^1$ of the Albanese). As far as I know, there is no general representation-theoretic argument of the type you envisage which would prove it; it is something deep and special about the particular representations appearing in etale cohomology.

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A remark on the (lack of) $p$-adic valued Haar measure: there can exist topologically irreducible continuous representations of pro-$p$ groups (e.g. $GL_n(\mathbb Z_p)$ on infinite-dimensional $p$-adic Banach spaces. This is pretty strong evidence that there is no simple analogue of Haar measure in the $p$-adic situation. (Remember that the existence of Haar measure means that this is impossible for representations of compact groups on real or complex spaces: topologically irreducible representations are necessarily finite-dimensional.)

Another remark, on semi-simplicity: It is conjectured that the representations of the absolute Galois group $G_K$ of $K$ occuring on the $p$-adic etale cohomology of a smooth proper variety over a number field $K$ is semi-simple. (This is a part of the so-called Tate conjecture.) As far as I know, this is very wide open other than in the abelian variety case (and hence in the case of $H^1$ in general, since $H^1$ can always be thought of as the $H^1$ of the Albanese).