Let $A$ and $B$ be $n\times n$ hermitian matrices satisfying $A \succeq \pm B$, where $\succeq$ denotes Loewner partial order. It is known that this is not sufficient to imply $A \succeq |B|$, where $|B| = \sqrt{B^\dagger B}$ denotes the absolute value/modulus of a matrix (if that claim were true, it would imply the inequalities suggested in https://mathoverflow.net/questions/173613/how-much-does-the-absolute-value-of-an-operator-behave-like-an-absolute-value and/or Does matrix modulus satisfy triangle inequality for Loewner order?, which are known to be false).
However, is it perhaps true that the weaker inequality $A \succeq \operatorname{Pos}(B)$ might hold? Here, $\operatorname{Pos}$ denotes the "positive part" of a hermitian operator, in the sense that since $B$ is hermitian it has a spectral decomposition $B = \sum_j \lambda_j P_j$ for eigenvalues $\lambda_j$ and projectors $P_j$, and we define $\operatorname{Pos}(B) = \sum_{j\in\mathcal{S}} \lambda_j P_j$ where $\mathcal{S}$ is the set of indices such that $\lambda_j \geq 0$.
(Edit: The above conjecture cannot be true; see the partial answer below. The following conjecture remains open.)
As a slightly weaker possibility, can we find some $k>1$ such that $k A \succeq |B|$ instead? (Note that $A \succeq \operatorname{Pos}(B)$ would imply $k=2$ suffices.)
Only a partial answer, but after a fair amount of numerical trial and error I seem to have found there's an example of $9\times9$ matrices such that $A\succeq \pm B$ but $A\nsucceq\operatorname{Pos}(B)$, hence disproving the first conjecture.