Let $G$ be a semi-simple linear algebraic group over $\mathbb{C}$ and $\rho\colon G \to \operatorname{GL}_{\mathbb{C}}(V)$ be a finite-dimesional irreducible rational representation of $G$.
Being a linear group, one can talk about the $\mathbb{Z}$-points of $G$: $G(\mathbb{Z}) = G\cap \operatorname{GL}_n(\mathbb{Z})$ if $G \leq \operatorname{GL}_n(\mathbb{C})$. As algebraic groups, we have the extension of scalars $G = G(\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{C}$ -- if I understand correctly this is true because we can recover $G(\mathbb{Z})$ by applying the construction of the Universal Chevalley Group to the Lie algebra of $G$ and then killing out a central subgroup determined by the root lattices.
Now, given this rational representation, can we always find a lattice $V_{\mathbb{Z}}$ such that $\rho$ is the extension of scalars of a rational representation $G(\mathbb{Z}) \to \operatorname{GL}_{\mathbb{Z}}(V_{\mathbb{Z}})$?
I'm not very familiar with the Chevalley constructions, but if I understand correctly one first builds special bases to complex semisimple Lie algebras to obtain a lattice called Chevalley bases, and then applying the exponential formulas to this base in order to find a linear algebraic group. One then checks that the elements in the Chevalley bases of the Lie algebra gets sent to integral elements in the algebraic group. Given the construction of the irreducible rational representations by Verma modules, I would suppose one could repeat this procedure to construct the map $G(\mathbb{Z}) \to \operatorname{GL}_{\mathbb{Z}}(V_{\mathbb{Z}})$, but I couldn't find a reference as I'm not sure I'm searching the right terms.