Strictification itself costs no Choice; choosing a skeleton can

The construction in Mac Lane strictification chooses nothing: from a given monoidal category $\mathcal{C}$ it builds the right-module endofunctor category ${\mathcal{C}}^{\mathrm{\prime }}$ and the functor $L:\mathcal{C}\to {\mathcal{C}}^{\mathrm{\prime }}$ directly. So the strictification theorem itself is a ZF statement.

Choice can enter through a separate operation: passing from an arbitrary category to a skeleton by selecting one representative from each isomorphism class. Being skeletal is a property and does not itself require a selection; it is the construction of a skeleton for a given category that may carry this set-theoretic cost. Consequently any argument that combines strictification with a chosen skeleton must account for that additional choice, while strictification alone does not incur it.