If I've understood the question correctly, the answer appears to be no again.

For example, let $\Psi : M_2 \rightarrow M_2$ such that
$$
\Psi\left(\left[\begin{array}{cc} a&b \\ c&d \end{array}\right]\right)
= \frac{2}{5}\left[\begin{array}{cc}\frac{3}{2}a+d &c \\ b & \frac{3}{2}d+a \end{array}\right].
$$
Consider that $\Psi$ is unital and invertible and
$$
\left[\begin{array}{cc}\Psi(E_{11}) & \Psi(E_{12}) \\ \Psi(E_{21}) & \Psi(E_{22})\end{array}\right]
= \frac{2}{5}\left[\begin{array}{cccc}\frac{3}{2}\\ &1&1\\&1&1\\&&&\frac{3}{2}\end{array}\right] \geq 0
$$
gives by Choi's theorem that $\Psi$ is completely positive.
Because it is invertible there is only one unique choice for a retract: $\Phi = \Psi^{-1}$. However for $\|\left[ \begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right] \| \leq 1$
$$
\left\|\Phi\left(\left[\begin{array}{cc} a&b \\ c&d \end{array}\right]\right)\right\|
= \left\|\frac{5}{2}\left[\begin{array}{cc} (6a-4d)/5 &c \\ b & (6d-4a)/5 \end{array}\right]\right\|
$$
$$= \frac{5}{2}\left\|\left[\begin{array}{cc} a & c \\ b & d \end{array}\right] + \left[\begin{array}{cc} (a-4d)/5 \\ & (d-4a)/5 \end{array}\right]\right\| \leq \frac{5}{2}(1 + 1) = 5
$$
but
$$
\Psi^{(2)}\left( \left[ \begin{array}{cccc} 1 \\ &&1 \\ &1\\ &&&1\end{array} \right]   \right)
= \frac{5}{2}\left[\begin{array}{cccc} 6/5& &&1 \\ &-4/5 \\ &&-4/5 \\ 1& &&6/5\end{array}\right] > 5
$$
which implies that $\|\Phi\|_{cb} > 5$. Therefore, $\|\Phi\| < \|\Phi\|_{cb}$.