You don't need Stinespring's theorem for this. How this could work (without details):
At first you get a map $\varphi\times \psi:A\odot B\to B(H), a\otimes b\mapsto \varphi(a)\psi (b)$, where $A\odot B$ is the $*$-algebraic tensor product of $A$ and $B$. To extend $\varphi\times \psi$ to a map on $A\otimes_{max}B$ you can do the following: Prove that this map is positive ( for this you need that the ranges of $\varphi$ and $\psi$ commute). Then for every positive linear functional $\eta :B(H)\to \mathbb{C}$ the composition $\eta \circ (\varphi\times \psi):A\odot B\to \mathbb{C}$ extends uniquely to a positive linear functional on $A\otimes_{max}B$. Conclude that there exists a well-defined map $$\Gamma: B(H)^*\to (A\otimes_{max}B)^*,\; \eta \mapsto \eta \circ (\varphi\times \psi),$$where $B(H)^*$ denotes the bounded linear functionals on $B(H)$. This map is bounded (use closed graph theorem for this), thus $\varphi\times \psi$ extends to a bounded map $\varphi\otimes_{max} \psi:A\otimes_{max}B\to B(H)$. To prove that this map is completely positive, fix an arbitrary $n\in\mathbb{N}$ and prove that $$(\varphi\otimes_{max} \psi)^{(n)}:M_n(A\otimes_{max}B)\to M_n(B(H)),\; (z_{ij})_{i,j}\mapsto (\varphi\otimes_{max} \psi (z_{ij}))_{ij}$$ is positive. To do this, note that the product of commuting positive elements is again positive.
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