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).
Just to turn my comment into an answer (let me know if more clarification is needed).
Since $G$ and $FG$ are connected we know that $G(k)=FG(k)=\{e\}$. Now, for any finite flat group scheme $H$ over $k$ we have that $\mathrm{Lie}(H)=\ker(H(k[\varepsilon])\to H(k))$ where $k[\varepsilon]\to k$ is the unique $k$-algebra map. In particular, if $H$ is connected then $\mathrm{Lie}(H)=H(k[\varepsilon])$. So, if $\mathrm{Frob}:G\to FG$ were injective then, in particular, the induced map $G(k[\varepsilon])\to FG(k[\varepsilon])$ is injective. Or, in other words, we'd have that the map $\mathrm{Lie}(G)\to\mathrm{Lie}(FG)$ is injective. But, it's not hard to see that the Frobenius map induces the zero map on Lie algebras (since $da^p=p da^{p-1}=0$ since the Lie algebra is a $k$-space).
Best Answer
Notations:
$$D_k=W(k)[F, V]/(Fa=a^pF, aV=Va^p, FV=VF=p \text{ for all } a\in W(k) )$$
$$W_k=\mathsf{Spec}(k[t_0, t_1, \cdots])$$
$$K_n:=\mathbf{Ker}(p^n\cdot id_{W_k})\subset W_k$$
$$W(p)_{k}=\varinjlim_n K_n$$
$${}_{n}W_m=\mathsf{Spec}(k[t_0, \cdots, t_{n-1}]/\big( t_0^{p^m}, \cdots, t_{n-1}^{p^m} \big))$$
Lemma:
Let $k$ be a perfect field and $f(x, y)\in W(k)[x,y]$ a polynomial with $deg_{y}f>0$. The formal group $G=\mathsf{Ker}(f(F, V):W(p)_k\to W(p)_k)$ is a $p$-divisible group with Dieudonné module $$ M(G)=\varprojlim_n D_{k}/(F^n, f(F, V)).$$
Proof:
Put $G_n=\mathsf{Ker}(f(F, V):K_n\to K_n)$, then $G=\varinjlim_{n} G_n$
(as $\{ K_n \}_{n\geq 0}$ form a filtered system, its injective limit preserves kernel).
We define $G_{n,m}$ by the following exact sequence $$0\to G_{n,m} \to {}_{n}W_m \xrightarrow{f(F, V)} {}_{n}W_m. $$ Taking the inverse limit (remark that inverse limit commutes with kernel) and we have $G_{n}=\varprojlim G_{n,m}$.
We know that $M({}_{n}W_m)\simeq D_{k}(F^n, V^m)$, and hence we have $$ M({}_{n}W_m)\xrightarrow{f(F, V)} M({}_{n}W_m) \to M(G_{n, m}) \to 0, $$ and hence $M(G_{n,m})= D_k/(F^n, V^m, f(F, V))$.
Then we have $$M(G_n)=\varinjlim_m M(G_{n,m})=\varinjlim_m D_k/(F^n, V^m, f(F, V))$$
$$= D_k/(F^n, f(F, V)).$$
Hence $$ M(G)=\varprojlim_n D_{k}/(F^n, f(F, V)).$$
This gives the Dieudonné module of $G$.
Conversely, by the equivalence of category given by Demazure (in page 71), we can construct a $p$-divisible group $G^{'}:=\varinjlim_n G_n^{'}$ where $G_n^{'}$ is defined to be the group scheme whose Dieudonné module is $M(G)/p^n$. Then we check that $G=G^{'}$.
(Here needs some details: We can see that $M(G_{n(r+s)}^{'})\twoheadrightarrow M(G_{n(r+s)}) \twoheadrightarrow M(G_n^{'})$ and hence $M(G)=M(G^{'})$, but then how to conclude? Shouldn't we have even $G^{'}_n=G_n$, as both should be $G[p^n]$?)
Example: When $k=\mathbf{F}_p$, and $f(F, V)=F^r-V^s$, we have $$ M(G)=\mathbf{Z}_p[F, V]/(FV-p, VF-p, F^r-V^s). $$