Let $G$ be a discrete group. Let $(A,\alpha)$ and $(B,\beta)$ be $G$-$C^*$-algebras and $\varphi: A \to B$ be $G$-equivariant and completely positive. All crossed products in this post are full (= universal).

I want to prove that there is a completely positive map $$\varphi \rtimes G: A \rtimes_\alpha G \to B \rtimes_\beta G$$ such that $(\varphi\rtimes G)(\sum_s a_s s) = \sum_s \varphi(a_s)s$.

I managed to prove the following using the $G$-equivariant Stinespring theorem:

Assume that $u: G\to B(H)$ is a unitary representation and $\sigma: A \to B(H)$ a completely positive map satisfying $\sigma(\alpha_g(a)) = u_g \sigma(a)u_g^*$ (i.e. $\sigma$ is $G$-equivariant where $B(H)$ has the $G$-action induced by $u$). Then there is a unique completely positive map $$\sigma \rtimes G: A \rtimes G \to B(H)$$ satisfying $\sigma \rtimes G(\sum_s a_s s) = \sum_s \sigma(a_s)u_s.$

I think I might be able to use this result to prove the result I want: Maybe the following works:

Let $(u,\pi)$ be a covariant representation of $B \rtimes G$ on $H$ where $\pi$ is chosen faithful. Then consider the composition $\sigma: A \to B(H)$ defined by $$A \stackrel{\varphi}\to B \stackrel{i}\to B \rtimes G \stackrel{\pi}\to B(H)$$ which is completely positive and satisfies $\sigma(\alpha_g(a)) = u_g \sigma(a)u_g^*$. Hence, by the above result, we obtain an induced map $$\sigma \rtimes G: A \rtimes G \to B(H).$$

If we can check that $(\sigma \rtimes G)(A \rtimes G) \subseteq \operatorname{Im}(\pi)$ then we can define $$\varphi \rtimes G(x) = \pi^{-1}(\sigma \rtimes G)(x) \in B \rtimes G$$ which would yield the desired extension. However, I don't think the above inclusion holds.