Positivity is not enough (complete positivity is).

Indeed, in $M_2(\mathbb{C})$, let $\Lambda\left(\left[\begin{array}{cc}a&b \\\\ c&d\end{array}\right]\right)=\left[\begin{array}{cc}a&0 \\\\ 0&a\end{array}\right]$. Then $\Lambda$ is positive and $\Lambda(I)=I$; but if $B=\left[\begin{array}{cc}1&0 \\\\ 0&0\end{array}\right]$ we have, for any $1\leq p<\infty$, $\|B\|_p=1$, $\|\Lambda(B)\|_p=2^{1/p}$.

For a completely positive map, Stinespring leads to an easy proof that the inequality holds for $p=2$. For other $p$ it looks trickier, but I think it might work too.

Positivity is not enough (complete positivity is).

Indeed, in $M_2(\mathbb{C})$, let $\Lambda\left(\left[\begin{array}{cc}a&b \\\\ c&d\end{array}\right]\right)=\left[\begin{array}{cc}a&0 \\\\ 0&a\end{array}\right]$. Then $\Lambda$ is positive and $\Lambda(I)=I$; but if $B=\left[\begin{array}{cc}1&0 \\\\ 0&0\end{array}\right]$ we have, for any $1\leq p<\infty$, $\|B\|_p=1$, $\|\Lambda(B)\|_p=2^{1/p}$.

For a unital completely positive map, Stinespring leads to an easy proof that the inequality holds for $p=2$. For other $p$ it looks trickier, but I think it might work too.