For a matrix $A$ with complex entries, the modulus $|A|=\sqrt{A^{\dagger}A}$ is a Hermitian positive semidefinite matrix, so we can consider the Loewner order $B\prec A$ iff $A-B$ is positive semidefinite.
Do we have then $$|A+B|\prec |A|+|B|,$$ for any pair of matrices $A$,$B$?
Is there a counterexample to this? Any hint to prove it otherwise?
It seems this is answered in the negative at https://mathoverflow.net/questions/173613/how-much-does-the-absolute-value-of-an-operator-behave-like-an-absolute-value (in particular, see the answer https://mathoverflow.net/a/279621 --- by choosing $A=p_1$ and $B=-p_2$, where $A,B$ are in your notation and $p_1,p_2$ are in the notation in the linked answer, you obtain an instance where $|A+B| = |p_1-p_2| \npreceq p_1 + p_2 = |A| + |B|$).