Do finite simplicial sets jointly detect isomorphisms in the homotopy category?

The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.

Nice question!

I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote

''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Z\subset X$ the restriction $f|_Z:Z\to Y$ is homotopic to a constant."

The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^{(1)}$ group associated to the automorphisms of the spaces.

This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.

A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^{(1)}$ groups. The intuition is that the homotopies needed to give the isomorphisms $$\mathcal{H}(K,X) \cong \mathcal{H}(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.

Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.

I hope this helps.