TITLE: Characterizing the rationalization of spaces.
QUESTION [5 upvotes]: In the category of rational spaces, loop spaces split as products of Eilenberg-Mac Lane
spaces and SUSPENSIONS split as wedges of (rational) spheres.  I wonder if anything of the following form is true:
(*)  Any functor $F$ from spaces to spaces which splits suspensions and loop spaces as above must factor through the rationalization.
EDIT 1:  Greg raises some fine questions, but I stand by my wording.  This is a question that arises from curiosity, not because I need it for anything, so I'd be happy with "anything like" the given statement.  
EDIT 2:  At least for simply-connected spaces, rationalization commutes with loop and suspension.  But, it seems to me that the power of the property is that the suspension
of any F-space splits and the loops of any F-space splits.  So I would go with:
the suspension of any rational space splits as a wedge of rational spheres and 
the loops of any rational space splits as a product of rational Eilenberg-Mac Lanes spaces.
Thus, we'd be looking for functors to some model-esque category with some relatively manageable list of objects whose products exhaust the homotopy types of loop spaces and whose wedges exhaust the homotopy types of suspensions.

REPLY [3 votes]: I think the following is a trivial counterexample, which may lead you to reflect about your question:
\begin{align*}
F\colon  Spaces & \longrightarrow Spaces\\\\
X&\;\mapsto\;\bigvee_{H_1(X,\mathbb{F}_2)}S^1
\end{align*}
This functor takes any space to a wedge of several circles, one circle for each element in the homology group ${H_1(X,{\mathbb{F}}_{2})}$. Such wedges are both suspensions and Eilenberg-MacLane spaces. Obviously this functor does not factor through rationalization, since there are spaces $X$ and $Y$ with $X\simeq _{\mathbb{Q}} Y$ but $|H_1(X,\mathbb{F}_2)|\neq |H_1(Y,\mathbb{F}_2)|$. 
Of course, you can replace $H_1$ with $H_n$ for any $n$ if you wish to work with simply connected spaces.