For $G$ a finite group, and algebraically closed field $k$ of finite characteristic coprime to $|G|$, we have that $k[G]$ is a semisimple algebra. This gives complete reducibility, and I vaguely recall the degrees of the irreps are the same as the irreps in characteristic 0.
How "good" is the bijection of the irreducible representations over $k$ and those over $\mathbb{C}$?
Is this bijection functorial with respect to the induction/restriction functors, (or at least on the level of isoclasses)?
Does the bijection extend to an isomorphism of the groethendieck rings of representations $R_k[G]$ and $R_\mathbb{C}[G]$?
You might as well assume that $k=\Bbb F_{p^k}$, a finite field, as all characteristic $p$ representations will be realisable over such a field. By the formalism of Brauer characters, such representations can be lifted to reps defined over the ring $R_p$, the ring of integers of the maximal unramified extension $\Bbb Q_p^{\text{ur}}$ of $\Bbb Q_p$. (You can realise $R_p$ as a ring of Witt vectors over $k$.) Likewise given a representation of $G$ on a vector space over $\Bbb Q_p^{\text{ur}}$, there's a $R_p$-lattice $\Lambda$ invariant under $G$, and reducing modulo $p$ gives a representation over $k$. All of this is functorial, and for details see Serre's book on representation theory.
To bring in $\Bbb C$ one needs an embedding $i:\Bbb Q_p^{\text{ur}}\to\Bbb C$. Once one has this one has a functorial correspondence between reps over $i:\Bbb Q_p^{\text{ur}}$ and over $\Bbb C$. But this embedding is not unique, and there is no canonical embedding. So the answer to your question is "yes, up to the choice of the embedding $i$."