Let $A$ be a unital subspace of a $C^*$-algebra. Then every contractive linear map $\varphi$ with $\varphi(1) = 1$ from $A$ into a commutative $C^*$-algebra is completely contractive.
Can anything similar be said for isometric maps, i.e., when is an isometry a complete isometry?
Indeed, if $\varphi$ is a unital isometry, then $\varphi^{-1}$ is unital and contractive on $\varphi(A)\subset B$. Let $f\in S(A)$ (that is, a state on $A$). Then $f\circ\varphi^{-1}$ is a unital and contractive functional on the operator system $\varphi(A)$. It follows that it is positive (the states on an operator system are precisely the unital contractive functionals). As $f$ is arbitrary, we deduce that $\varphi^{-1}$ is positive; but with abelian domain, so it is completely positive. As it is also unital, it is completely contractive.
As mentioned in the question, since $\varphi$ a unital contaction, it is completely contractive. So both $\varphi$ and $\varphi^{-1}$ are completely contractive, which shows that $\varphi$ is a completle isometry.