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Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked. Ah well... I think Brian's comments above essentially answer the question, though. See below for example for finite $k$.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$. This isn't special to tori: e.g. if $G = \operatorname{SL}(V)_{/\mathbf{F}_q}$ then $V$ and the first Frobenius twist $V^{[1]}$ are distinct simple algebraic $G$-modules, but they give isomorphic representations of $G(\mathbf{F}_q)$.

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$. 

This isn't special to tori: e.g. if $G = \operatorname{SL}(V)_{/\mathbf{F}_p}$ then $V$ and the first Frobenius twist $V^{[1]}$ (defined by twisting the action of $\operatorname{SL}(V)$ on $V$ by the Frobenius map) are distinct simple algebraic $G$-modules, but they give isomorphic representations of $G(\mathbf{F}_p)$.

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked. Ah well... I think Brian's comments above essentially answer the question, though. See below for example for finite $k$.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$.

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked. Ah well... I think Brian's comments above essentially answer the question, though. See below for example for finite $k$.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$. This isn't special to tori: e.g. if $G = \operatorname{SL}(V)_{/\mathbf{F}_q}$ then $V$ and the first Frobenius twist $V^{[1]}$ are distinct simple algebraic $G$-modules, but they give isomorphic representations of $G(\mathbf{F}_q)$.

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked. Ah well... I think Brian's comments above essentially answer the question, though. See below for example for finite $k$.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not faithful. Indeed, if $k = \mathbf{F}_q$ then $k_1$ and $k_q$ are different representations of $T$ but are the same when restricted to $T(k)$.

Edit Sorry: I just realized that I conflated "essentially surjective" with "full" in my head this afternoon. So this is mostly an answer to a question that wasn't asked. Ah well... I think Brian's comments above essentially answer the question, though. See below for example for finite $k$.

The functor is not essentially surjective, already when $G = T$ is a split torus.

Take for example the 1 dimensional torus $T = \mathbf{G}_m$; in this case, the 1 dimensional algebraic representations of $T$ are parametrized by $m \in \mathbf{Z} = X^*(T)$; to an $m$ corresponds the 1 dimensional representation $k_m = k$ on which $T$ acts with weight $1$ (so in particular an element $t$ of $T(k) = k^\times$ acts by multiplication with $t^m$).

Now let $V$ be the representation $k_1$, let $\sigma$ be a non-trivial automorphism of the field $k$, and let $\ ^\sigma V$ be the representation of $T(k) = k^\times$ obtained from $V$ by "twisting with $\sigma$"; thus an element $t \in T(k) = k^\times$ acts by multiplication with $\sigma(t)$.

If $k$ has characteristic 0, or if $\sigma$ is not a power of the Frobenius automorphism, the representation $\ ^\sigma V$ is not isomorphic as $T(k)$-representation to $k_m$ for any $m$. Thus, the functor isn't essentially surjective.

When $k$ is finite, this construction shows that the indicated functor is not full. Indeed, if $k = \mathbf{F}_q$ then $\operatorname{Hom}_T(k_1,k_q) = 0$ but $\operatorname{Hom}_{T(k)}(k_1,k_q) = k$ since $k_1$ and $k_q$ are isomorphic when restricted to $T(k)$.