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).
You are right that $\operatorname{Frob}_v$ is well-defined only up to a conjugation, but you don't really need to choose an $L$. This is because the Frobenius for different $L$ can be packed together (as a projective limit) to $\overline K$. Let me explain this below.
Let $K$ be a global field (number field or function field). If $v$ is a finite place of $K$, then we may form the $v$-adic completion $K_v$, so that $K$ is canonically embedded in $K_v$.
Let $w$ be any lift of $v$ to $\overline K$. The choice of $w$ is equivalent to choosing an embedding $\overline K \hookrightarrow \overline {K_v}$. This induces an injection of groups $\operatorname{Gal}(\overline {K_v}/K_v) \hookrightarrow \operatorname{Gal}(\overline K / K)$, whose image is the decomposition group $D_{w/v}$. Note that two different choices of embeddings $\overline K \hookrightarrow \overline {K_v}$ differ by an automorphism of $\overline K$, hence the two resulting decomposition groups differ by a conjugation.
Now in the case of a local field $K_v$ with residue field $\Bbb F_q$, the residue field of $\overline {K_v}$ is simply $\overline{ \Bbb F_q}$. Thus any automorphism $\phi\in \operatorname{Gal}(\overline {K_v}/K_v)$ induces an automorphism $\psi \in \operatorname{Gal}(\overline {\Bbb F_q}/\Bbb F_q)$.
This gives rise to a surjective homomorphism of groups $\tau: \operatorname{Gal}(\overline {K_v}/K_v) \rightarrow \operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$.
We know that $\operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$ is canonically isomorphic to $\widehat{\Bbb Z}$, the profinite completion of $\Bbb Z$, and we call the inverse image of $1\in \widehat{\Bbb Z}$ in $\operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$ the "Frobenius", denoted $\operatorname{Frob}_v$.
The inverse image $\tau^{-1}(\operatorname{Frob}_v) \subseteq \operatorname{Gal}(\overline {K_v}/K_v)$ is a coset of the kernel $I_v$, the inertia group. But by abuse of notation, it is also called "the" Frobenius, and by even more abuse of notation, any element in this coset is called "the" Frobenius.
To add even more confusion, there is a difference of arithmetic vs geometric Frobenius, which is we choose the inverse image of $-1\in \widehat{\Bbb Z}$ above.
Back in the global situation, we have the embedding $\operatorname{Gal}(\overline{K_v}/K_v)\hookrightarrow \operatorname{Gal}(\overline K/K)$ and the image of a Frobenius is again called a "Frobenius".
In view of this unfortunate situation, one should always pay attention to what is meant by "the" Frobenius.
Best Answer
This is really a local question: you may as well assume $X$ is defined over a local field $L$, it doesn't matter whether it comes from a number field.
The point is that if $S = \operatorname{Spec} O_L$ and $\pi: \mathfrak{X} \to S$ is the (smooth proper) structure map, then for any locally constant torsion sheaf $\mathcal{F}$ on $\mathfrak{X}$ whose order is invertible on $S$, there is a locally constant sheaf $R^i\pi_\star\mathcal{F}$ on $S$ whose fibre at any geometric point $\bar{x}$ is $H^i(\mathfrak{X}_{\bar{x}}, \mathcal{F}_{\bar{x}})$; see Corollary VI.4.2 of Milne's "Lectures on Etale Cohomology" for the details. Since $S$ is connected, we get an isomorphism between the fibres at the closed point of $S$ and the generic point. That is, $$H^i(X, \mathcal{F}) = H^i(X_0, \mathcal{F}_0)$$ where $X_0$ is the special fibre of $\mathfrak{X}$. This isomorphism is Galois-equivariant; but the inertia subgroup acts trivially on $X_0$. Hence $H^i(X, \mathcal{F})$ is unramified.