A recipe for a counterexample: 

Let $(X,e,m:X\wedge X\to X)$ be a monoid object in the homotopy category of pointed spaces $h\mathcal{S}_\ast$ (that is, an H-monoid), where $\wedge$ denotes the smash product in the pointed homotopy category.  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 $$\mu:X\times X \to X\wedge X \xrightarrow{m} X.$$ 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 $A_3$-algebra structure on $X$, and there exist many examples (none of which I know!) of $A_3$-structures that don't admit a lift to an $A_\infty$ structure.  

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 $(X,e,m)$.