Please correct me if I am wrong, but I really don't see where we need the proper assumption to get a Galois action.
So let $X/k$ be any scheme, $\overline{k}$ a separable (or algebraic) closure of $k$, and $\mathcal{F}$ be any sheaf on $X$. Write $X_\overline{k}$ for the base change of $X$, $p:X_\overline{k}\rightarrow X$ the induced morphism. Write also $\mathcal{F}_\overline{k}=p^*\mathcal{F}$ for the pullback of $\mathcal{F}$ to $X_\overline{k}$.
For $g\in\mathrm{Gal}(\overline{k}/k)$, we have an induce morphism $g:X_\overline{k}\rightarrow X_\overline{k}$, which induce a pullback $g^*\mathcal{F}_\overline{k}$. I claim that there is a canonical isomorphism $g^*\mathcal{F}_\overline{k}\simeq\mathcal{F}_\overline{k}$. This is because $\mathcal{F}$ comes from $X$. Indeed : $g^*\mathcal{F}_\overline{k}=g^*p^*\mathcal{F}=p^*\mathcal{F}=\mathcal{F}_\overline{k}$.
Hence we have an induced action on cohomology :
$$ H^i(X_\overline{k},\mathcal{F}_\overline{k})\rightarrow H^i(X_\overline{k},g^*\mathcal{F}_\overline{k})\simeq H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
which is the action you are looking for.
Now there is also the approach of Alex Youcis : if $f:X\rightarrow\operatorname{Spec}k$ is the structural morphism, $R^if_*\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$ hence a set equipped with a continuous discrete $\operatorname{Gal}(\overline{k}/k)$-set. Unless I'm mistaken, I don't think we need to add any assumption for the following claim : the underlying $\operatorname{Gal}(\overline{k}/k)$-set is exactly $H^i(X_\overline{k},\mathcal{F}_\overline{k})$ with the above action. (This prove in particular that the above action is continuous).
This follows from the following continuity result
$$\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
where $k'$ runs through the finite Galois extensions of $k$ in $\overline{k}$ and the trivial base change $(R^if_*\mathcal{F})_{k'}=R^if_*\mathcal{F}_{k'}$.
EDIT : Let me expand a bit on the last assertion. First, let us show that the stalk of $R^nf_*\mathcal{F}$ at a geometric point $\operatorname{Spec}\overline{k}$ is indeed $H^n(X_\overline{k},\mathcal{F}_\overline{k})$.
By definition, the stalk of $R^nf_*\mathcal{F}$ is $\varinjlim_{k'}R^nf_*\mathcal{F}(k')$ where the limit is taken over all the finite extension of $k'$ inside $\overline{k}$. Now recall that $R^nf_*\mathcal{F}$ is the sheaf associated to the presheaf $k'\mapsto H^n(X_{k'},\mathcal{F}_{k'})$, and since the stalk of a presheaf is the same as its associated sheaf, we get $(R^nf_*\mathcal{F})_\overline{k}=\varinjlim_{k'}H^n(X_{k'},\mathcal{F}_{k'})=H^n(X_\overline{k},\mathcal{F}_\overline{k})$, the last equality is from a limit argument (we should add $X$ quasi-compact and quasi-separated here).
Recall that the equivalence between sheaves on $(\operatorname{Spec}k)_{ét}$ and $\operatorname{Gal}(\overline{k}/k)$-sets is the following : if $\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$, then $\mathcal{F}_\overline{k}=\varinjlim\mathcal{F}(k')$ is a $\operatorname{Gal}(\overline{k}/k)$-set. The action of $\sigma\in\operatorname{Gal}(\overline{k}/k)=\varprojlim\operatorname{Gal}(k'/k)$ on $\mathcal{F}_\overline{k}$ is induced by the compatibles actions on the $\mathcal{F}(k')$ (check that is indeed compatible).
Now you should convinced yourself that this is indeed the pull-back to $\overline{k}$ and that this action is the same as the induced action by functoriality (that is very first one I wrote). Do the same with $R^if_*$ and you will get the compatibility of the two actions in general.
Exercise : Take $k=\mathbb{R}$ and $\mathcal{F}=\mu_4$ the sheaf $A\mapsto\{x\in A, x^4=1\}$ and compute the action of the stalk (there is no trap here).
Best Answer
No. I think you might be misinterpretting what 'locally trivial fibration' means.
Namely, what Ehresmann's theorem implies is the following:
One cannot replace diffeomorphic by 'biholomorphic' or, if $X$ and $Y$ are algebraic and the $U_i$ is a Zariski cover, 'algebraic'.
Here's a silly counterexample to your literal claim.
Consider $D(\Delta)\subseteq \mathbb{A}^2_k=\mathrm{Spec}(k[a,b])$ where, for simplicity, $k$ is an algebraically closed field of characteristic greater than $3$ where
$$\Delta(a,b)=-16(4a^3+27b^2)$$
We then have the universal family of Weierstrass equations living over $D(\Delta)$
$$\mathcal{W}^\text{univ}=V(y^3-axz^2-bz^3)\subseteq \mathbb{P}^2_k\times D(\Delta)$$
We then note that
$$\mathcal{W}^\text{univ}\to D(\Delta)$$
is a family of elliptic curves and so, in particular, smooth proper and surjective. Moreover, each $\mathcal{W}^\text{univ}$ and $D(\Delta)$ is smooth.
That said it's not an etale local fibration since this, in particular, would imply that the fibers of $\mathcal{W}^\text{univ}\to D(\Delta)$ are Zariski locally isomorphic. Indeed, let $\{U_i\to D(\Delta)\}$ be an etale cover such that $\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\cong U_i\times B_i$ for some $k$-space $B_i$. Then, in particular we see that for all $x\in V_i:=\text{im}(U_i\to D(\Delta))$ (this is an open set) we have that
$$\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\times_{V_i}x\cong \bigsqcup_{y\mapsto x}B_i$$
which since $k(x)\to k(y)$ is an isomorphism implies that $\mathcal{W}^\text{univ}_x\cong B_i$. Thus, $\mathcal{W}^\text{univ}_x\cong B_i$ for all $x\in V_i$.
Note that we have an algebraic map $j:D(\Delta)\to \mathbb{A}^1_k$, the $j$-function map, which sends, in particular, $x$ to the $j$-invariant $\mathcal{W}^\text{univ}_x$. Since $\mathcal{W}^\text{univ}_x$ is locally isomorphic this implies that $j$ is locally constant which implies, since $D(\Delta)$ is connected, that $j$ is constant and thus $\mathcal{W}^\text{univ}_x$ is actually constant for $x$ in $D(\Delta)$.
Of course, this is ludicrous since $\mathcal{W}^\text{univ}_x$ contains isomorphism classes of every elliptic curve over $k$.
Of course, it is true that the fibration $W^\text{univ}\to D(\Delta)$, when $k=\mathbb{C}$, is analytically locally on the target DIFFEOMORPHIC to the trivial fiber bundle. The operative point being that any two elliptic curves are diffeomorphic but are rarely isomorphic/biholomorphic.