A recipe for a counterexample:
Let $(X,e:\Delta^0\to X,m:X\times X\to X)$ be monoid object in the homotopy category of spaces $h\mathcal{S}$ (that is, an H-monoid). Note that this is a property of the triple. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$m:X\times X \to X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.
In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_2$-algebra structure on $X$ with the added stipulation that there exists an $A_3$-structure extending the $A_2$-structure, and there exist many examples (none of which I know!) of $A_2$-structures admitting (possibly many) extensions to an $A_3$ structure, none of which extend to an $A_\infty$-structure. Unfortunately, even having asked some experts, it's not so easy to come up with an example of such a space.
But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.
So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.
Edit: Kevin Carlson's comment below gives an example of such spaces from Stasheff and Adams and a source, so check it out!