The answer is "almost no" - the wavelength of the photon is virtually unchanged (in the initial rest frame of the mirror, the "lab frame"). Because the mirror is much more "massive" than the photon, it serves as a "momentum sink" and picks up almost no energy.
The best way to develop an intuition for this is to consider a collision between two balls: one lighter (with mass $m$) and initially moving (at velocity $v_1$) and one more massive (with mass $M$) and initially at rest. After scattering, the lighter ball leaves the scene at velocity $v_3$ and the more massive ball leaves the scene at velocity $v_4$.
Set $v_2 = 0$ in the following worked-out example (see page 3): http://www.its.caltech.edu/~teinav/Lectures/Ph%201a/Lecture%207%20-%202017-10-19.pdf
We obtain $v_3 = \frac{(M-m)v_1}{M+m}$ and $v_4 = \frac{2mv_1}{M+m}$
In the limit that $M >> m$ the fraction of the initial kinetic energy picked up by the massive object goes to zero, but it acquires twice as much momentum (and in the opposite direction) as the lighter object initially had. Thusly can momentum be transferred but (almost) no energy.
NOTE - In the "center of mass frame" the wavelength will be completely identical, but I believe it is the "lab frame" that supplies the intuition you seek. In the center of mass frame the momentum just changes sign and your equation should really be $|p| = E/c$, which admits a sign change of $p$ while conserving $E$. This is why there is a $2$ in the equation for $v_4$ - flipping the sign of momentum imparts double the initial momentum to the mirror.