I don't think this is literally true. For example, suppose that $G$ is a finite cyclic group of order 2 generated by $s$, suppose that $L$ is an non-trivial 2-torsion invertible $A$-module over a Dedekind ring $A$. Then $L^{\oplus 2}$ is isomorphic to $A^{\oplus 2}$; you can let $G$ act on $L^{\oplus 2}$ with the rule $s(a,b) = (a, -b)$. This gives a family of representations that does not come from $\mathbb C$. You can give a similar example in which $G$ is a torus.
On the other hand, what you want is true Zariski-locally; would that be enough for your needs?
[Edit] Let me add a proof that the statement is Zariski-locally true. The key point is the following: given a locally free sheaf on a scheme $X$ with an action of $G$, the subsheaf of invariants is also locally free. This is easy: if $X = \mathop{\rm Spec} A$ and $E$ corresponds to a projectve $A$-module $M$, then $M^G$ is a direct summand of $M = M^G \oplus M_G$, because $G$ is linearly reductive (and $M$ is a locally finite representation of $G$, this is standard); and $M_G$ is also an $A$-submodule, so $M^G$ is a projective $A$-module. Now, let $E$ be a locally free sheaf with an action of $G$. Let $p$ be a rational point of $X$, and $V$ be the representation of $G$ appearing as the fiber $E(p)$ of $E$ at $p$. Consider the locally free sheaf $Hom_{\mathcal O_X}(V \otimes \mathcal O_X, E)$; the sheaf of invariants, that is, the sheaf of $G$-equivariant homomorphisms $V \otimes \mathcal O_X \to E$, is locally free. At the point $p$ we have a section of its fiber, given by the tautological isomorphism $V \simeq E(p)$; since the sheaf is locally free, this extends to a section in a neighborhood. By further restricting the neighborhood, this sectioin will be an isomorphism, and this concludes the proof.