The answer to your example question is "No".
Let $G$ be isomorphic to $\widehat{\mathbf{Z}}$, and $\phi$ the topological generator of $G$ (corresponding to $1 \in \widehat{\mathbf{Z}}$). Then one computes $H^0(G, M) = M^{\phi = 1}$ and $H^1(G, M) = M / (\phi - 1)M$, where $\phi$ is the generator of $G$.
Now $M \cong \mathbf{Z}_p$ be the trivial one-dimensional representation, and let $M'$ be the unramified representation with Frobenius acting as $1 + p$. Then $H^1(G, M) = M = \mathbf{Z}_p$, whereas $H^1(G, M') = \mathbf{F}_p$.
(In this case, we do at least have $H^1(G, M) \otimes \mathbf{F}_p = H^1(G, M') \otimes \mathbf{F}_p$. But even this is just a small-number coincidence, coming from the fact that the $H^2$'s vanish, and will fail if you take more complicated groups.)
What is true is that if $M \cong M' \bmod p$ (and $G$ has reasonable finiteness properties), then the cohomology complexes $R\Gamma(G, M)$ and $R\Gamma(G, M')$ are "congruent mod $p$ in the derived category", in the sense that $R\Gamma(G, M) \otimes^{\mathbf{L}} \mathbf{F}_p = R\Gamma(G, M') \otimes^{\mathbf{L}} \mathbf{F}_p = R\Gamma(G, \overline{M})$ (where $\overline{M}$ is the common residual representation). But much can go wrong in translating that into a concrete statement about cohomology groups.
The situation for $H^1_f$ is worse – far, far worse – because in general there is no sensible intrinsic definition of $H^1_f$ of a mod $p$ representation. (A large part of the machinery of Iwasawa theory is designed to measure, control, and in some cases cleverly exploit, the discrepancy between $H^1_f$'s of congruent representations).