Assume two $CW$ complexes $X,Y$ give two functors $h_X=[,X], h_Y=[,Y]$ on the homotopy category of $CW$ complexes whose restrictions to the full subcategory of finite $CW$ complexes are naturally equivalent. Does this imply that the two functors are naturally equivalent (i.e. that X, Y are homotopy equivalent)?

3$\begingroup$ The main question is if natural equivalence of functors is represented by a map $X\to Y$. If so, then the spaces are weakly equivalent because this is tested on spheres, which are finite complexes. By Whitehead's theorem, they are also homotopy equivalent. Maybe the representability of the natural equivalence is discussed in Brown's paper? $\endgroup$ – Matthias Wendt Oct 3 '14 at 6:54
The answer depends on which category of CWcomplexes you have in mind. It is true in the category of based connected CWcomplexes simply by the Whitehead's Theorem.
EDIT: The concern raised by Matthias in the comment above is addressed in Brown's paper (Theorem 2.8 of Abstract Homotopy Theory). This is only proven in the based case, but even if it works in the unbased case too, then the following still applies.
However, dropping any of these assumptions makes the answer negative. In fact, Heller proves in Corollary 2.3 of On the representability of homotopy functors that there is no set of CWcomplexes such that functors represented by them detect weak equivalences. (This is closely related to the failure of Brown's representability in the unbased case, discussed here based on another paper by Freyd and Heller.)
In particular, finite CWcomplexes do not suffice since a map inducing bijections on homotopy classes of maps out of CWcomplexes might not be a $\pi_1$isomorphism. In general, it will only induce a bijection on conjugacy classes of $\pi_1$s. On the other hand, this answer shows that this works for maps between CWcomplexes with finitely generated fundamental groups.

1$\begingroup$ Thanks for an interesting answer. Could you please comment on why this is true in the category of based connected CW complexes? Note that I don't assume the existence of a map $X \to Y$ inducing the equivalence of the restriction to finite complexes. $\endgroup$ – user58951 Oct 3 '14 at 14:38

$\begingroup$ That's what my edit addresses. Brown proves in Thm. 2.8 that the space representing a functor on finite based CWcomplexes is determined up to weak equivalence. $\endgroup$ – Karol Szumiło Oct 3 '14 at 15:40