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Let me expand on Ariyan's 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 written as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.

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.

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.

Let me expand on Ariyan's answer; I write $\operatorname{H}^{2i}_{\operatorname{alg}}$ for the image 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 $\mathcal{N}^G$, where $\mathcal{N}$ is the kernel of the cycle class map. 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 written as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.

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 $$\delta : (\operatorname{H}^{2i}_{\operatorname{alg}})^G \rightarrow \operatorname{H}^1(k,\mathcal{N}).$$ Unfortunately I am not a geometer, so I don't have any idea whether $\delta$ is or is not always trivial.

Let me expand on Ariyan's 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 written as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.

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.

To amplify on Ariyan's answer, if $(\operatorname{CH}^i)^G \rightarrow (\operatorname{H}^{2i})^G$ 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 $\mathcal{N}^G$, where $\mathcal{N}$ is the kernel of the cycle class map. 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 written as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.

Therefore, the only other obstruction is the (possible?) lack of surjectivity of $(\operatorname{CH}^i)^G \rightarrow (\operatorname{H}^{2i})^G$, which may be measured by the non-vanishing of the coboundary map $$\delta : (\operatorname{H}^{2i})^G \rightarrow \operatorname{H}^1(k,\mathcal{N}).$$ Unfortunately I am not a geometer, so I don't have any idea whether $\delta$ is or is not always trivial.

Let me expand on Ariyan's answer; I write $\operatorname{H}^{2i}_{\operatorname{alg}}$ for the image 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 $\mathcal{N}^G$, where $\mathcal{N}$ is the kernel of the cycle class map. 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 written as the sum of a $G$-invariant cycle and an element of $\mathcal{D}$.

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 $$\delta : (\operatorname{H}^{2i}_{\operatorname{alg}})^G \rightarrow \operatorname{H}^1(k,\mathcal{N}).$$ Unfortunately I am not a geometer, so I don't have any idea whether $\delta$ is or is not always trivial.