TheUnfortunately, the answer is no.
Suppose $\Phi : M_4 \rightarrow M_2$ and $\Psi : M_2 \rightarrow M_4$ are given by $$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) = A + 10B^T \ \ \textrm{and} \ \ \Psi(A) = \left[\begin{array}{cc} A&0\\ 0&A\end{array}\right] $$ where $A,B,C,D\in M_2$. Then $\Phi$ is unital, completely bounded, $\Psi$ is unital, completely positive and $\Phi\circ\Psi = I_{M_2}$ but $$\|\Phi\| \leq 11 \ \ \textrm{and} \ \ \|\Phi\|_{\rm cb} \geq 20$$ since the transpose map is cb-norm 2.
There is also a negative answer if you look for $\Phi$ being left-invertible by a ucp map:
In this case a counterexample is given by $\Phi : M_2 \rightarrow M_4$ and $\Psi : M_4 \rightarrow M_2$ defined by $$ \Phi(A) = \left[\begin{array}{cc}A & 0 \\ 0 & A^T\end{array}\right]\ \ \textrm{and} \ \ \Psi\left(\left[\begin{array}{cc} A & B \\ C & D\end{array}\right]\right) = A $$ where $A,B,C,D \in M_2$. Then $\Phi$ is ucb, $\Psi$ is ucp and $\Psi\circ\Phi = I_{M_2}$ but $$ \|\Phi\| = 1 < 2 = \|\Phi\|_{\rm cb}. $$