A graph argument settles this issue very nicely, as follows.

Consider a linear algebraic group $G$ over a field $k$ of characteristic 0, and let $\mathfrak{g}$ be its Lie algebra. For a finite-dimensional $k$-vector space $V$, let $f:\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ be a representation. Provided that $G$ is connected, since ${\rm{char}}(k) = 0$ clearly there is at most one $k$-homomorphism $\rho:G \rightarrow {\rm{GL}}(V)$ such that ${\rm{Lie}}(\rho) = f$, so assuming $G$ is connected we seek conditions under which such a $\rho$ always exists.

Let $\mathfrak{h} \subset \mathfrak{g} \times \mathfrak{gl}(V)$ be the graph of $f$. Clearly if $\rho$ is to exist and the $k$-subgroup $H \subset G \times {\rm{GL}}(V)$ is its graph then $H$ is connected and $\mathfrak{h} = {\rm{Lie}}(H)$, so $H$ is uniquely determined (as a $k$-subgroup of $G \times {\rm{GL}}(V)$) since ${\rm{char}}(k) = 0$. So we seek conditions under which the Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g} \times \mathfrak{gl}(V)$ "exponentiates" to a connected closed $k$-subgroup $H$ (and then we need conditions to ensure ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism).

Assume $G$ is its own derived group, so $\mathfrak{g}$ is its own derived subalgebra (as ${\rm{char}}(k)=0$) and hence the same for $\mathfrak{h}$. Now comes the crucial point: it is a general fact over fields $k$ of char. 0 (see Cor. 7.9 in Ch. II of Borel's textbook on algebraic groups) that the derived subalgebra of any Lie subalgebra of a linear algebraic group over $k$ "exponentiates" to a connected closed $k$-subgroup. So $\mathfrak{h} = {\rm{Lie}}(H)$ for a unique connected closed $k$-subgroup $H \subset G \times {\rm{GL}}(V)$. The necessary and sufficient condition for $f$ to arise from some $\rho$ is that ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism.

This projection has Lie algebra map $\mathfrak{h} \rightarrow \mathfrak{g}$ that is the analogous projection which is visibly an isomorphism (due to the definition of $\mathfrak{h}$ as the graph of $f$), so $H \rightarrow G$ is an isogeny. As such, its kernel is etale (since ${\rm{char}}(k) = 0$) and hence central (since $H$ is connected), so it is a finite central $k$-subgroup of $H$. Thus, we just need that $G$ admits no nontrivial isogenous (smooth) connected central extension. This is automatic when $G$ is assumed to be semisimple and simply connected (in the sense of algebraic groups).

So we win whenever $G$ is a connected semisimple $k$-group that is simply connected. We also win whenever $G$ is a unipotent $k$-group (by entirely different arguments), but presumably you're not interested in that case.