Let $X$ be a smooth projective variety over a finite field $k$, $G=Gal(\bar{k}/k)$ and $\Gamma\in CH^i(\bar{X})$ such that:
Do we necessary have $\Gamma \in CH^i(\bar{X})^G$? What might be the obstructions?
Let me try to generalize Ariyan's answer, which already points in the right direction but does not give a completely general answer. I write $\operatorname{H}^{2i}_{\operatorname{alg}}$ for the image of the cycle class map, so that we have a short exact sequence $$ 0 \longrightarrow \mathcal{N} \longrightarrow \operatorname{CH}^i \stackrel{\operatorname{cl}}{\longrightarrow} \operatorname{H}^{2i}_{\operatorname{alg}} \longrightarrow 0, $$ where $\mathcal{N}$ is the kernel of the cycle class map.
If $(\operatorname{CH}^i)^G \rightarrow (\operatorname{H}^{2i}_{\operatorname{alg}})^G$ is surjective (this is the case for instance if $i=1$, because then the kernel of the cycle class map is an abelian variety over the finite field $k$, whose first Galois cohomology vanishes by a classical result of Lang) then the only obstruction consists of (non-$G$-invariant) elements in the divisible hull $\mathcal{D}$ of $\mathcal{N}^G$. In particular, torsion elements of $\mathcal{N}$ are contained in $\mathcal{D}$.
Indeed, under these assumptions there exists a Galois-invariant cycle $\Gamma'$ with $\operatorname{cl}(\Gamma') = \operatorname{cl}(\Gamma)$, so that $\Delta := \Gamma' - \Gamma$ is in $\mathcal{N}$ and $N\cdot\Delta$ is $G$-invariant, with $N$ as in your post, hence $\Gamma$ can be decomposed as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.
In fact, this decomposition exists for every $\Gamma$ whose image coincides with that of some $G$-invariant cycle. Therefore, the only other obstruction is the (possible?) lack of surjectivity of $(\operatorname{CH}^i)^G \rightarrow (\operatorname{H}^{2i}_{\operatorname{alg}})^G$, which may be measured by the non-vanishing of the coboundary map $$\delta : (\operatorname{H}^{2i}_{\operatorname{alg}})^G \rightarrow \operatorname{H}^1(G,\mathcal{N}).$$ Unfortunately I am not a geometer, so I don't have any idea whether $\delta$ is or is not always trivial.
Ulrich beat me to it, but let me answer your question as well and elaborate on his comment.
One obstruction is the possible presence of torsion in the Chow ring.
Let $E$ be an elliptic curve over $\mathbb Q$ and let $\overline E$ be the base-change to $\overline{\mathbb Q}$. Take $i=1$ (in your notation) and consider a torsion element $\Gamma$ of $CH^1(\overline E) = \mathrm{Pic}(\overline E)$. The cycle class map $CH^1(\overline E) \to H^2(\overline E_{et},\mathbb Z_\ell(1))\cong \mathbb Z_\ell(1) $ is given by the degree map.
If $\Gamma$ has degree zero (i.e. lies in $\mathrm{Pic}^0(\overline E)$), then $cl(\Gamma)$ is zero and is clearly Galois invariant.
Since $\Gamma$ is torsion (by assumption), there exists some integer $N$ such that $N\Gamma$ is Galois invariant (take $N$ to be the order of $\Gamma$ in the Picard group).
Thus, to answer your question negatively it suffices to take a torsion line bundle in $\mathrm{Pic}^0(\overline E)$ which isn't Galois invariant. As the torsion group of $E(\mathbb Q)$ is finite of some order $o$, it suffices to take $\Gamma$ of order $>o$. (In Ulrich's words, take $p$ a point of order $>o$ and let $\Gamma = [p] - [o]$.)
