The actual work being done here is by the Yoneda lemma. Brown's representability theorem tells you that these spaces represent the cohomology theory, turning natural transformations into morphisms is done by Yoneda.
That said, uniqueness holds.
Of course, one has to be a little bit more careful about what one means by "natural transformation" in this case. Natural transformation of what? If your natural transformation is of the whole cohomology theory in one go (aka a stable operation) then, indeed, you get a morphism of spectra. If your natural transformation acts on one particular level (aka an unstable operation) then you get a morphism of spaces.
Where you do not get uniqueness is if you have a family of unstable operations (aka a family of morphisms of spaces) which look as if they fit together to give a stable operation. You can get "phantom" morphisms, and there's a "$\lim^1$" term that controls this. A nice place to read about all of this is the papers by Boardman and Boardman, Johnson, and Wilson on stable and unstable cohomology operations (Handbook of algebraic topology, also available via Steve Wilson's homepage).
Update: To expand on that last point (I was deliberately vague because I didn't have Boardman's paper in front of me and couldn't remember which way the maps went): a stable operation (morphism of spectra) defines a compatible (under suspension) family of unstable operations (morphisms of spaces, indeed, morphisms of infinite loop spaces) but this assignment may not be injective (it is always surjective). There is a short exact sequence (attributed to Milnor, and (9.7) in Boardman's paper):
$$ 0 \to \lim_n{}^1 E^{k-1}(\underline{E}_n,o) \to E^k(E,o) \to \lim_n E^k(\underline{E}_n,o) \to 0 $$
The "o" means "pointed" and $\underline{E}_n$ is the $n$th component in the spectrum $E$. So if the $\lim^1$ term vanishes, you get an isomorphism (whence injectivity) but if not, then there may be "phantom" stable operations that are non-trivial but can't be detected in the unstable realm.