I'll say $K$ has "finitely generated" homotopy groups if there is a finite wedge of spheres $W = \bigvee S^{n_i}$ and a map $f: W\to K$ which induces a surjection on $\pi_*$.
It seems likely that this is equivalent to saying that $\pi_*(K)$ is finitely generated as a $\Pi$-algebra, but I'm not sure of this.
Write $\mathcal{F}$ for the collection of all finite complexes with finitely generated homotopy groups. Clearly every finite wedge of spheres is in $\mathcal{F}$, as are the finite products of spheres and the projective spaces. $\mathcal{F}$ is closed under products.
QUESTION 1: I wonder if it is conceivable that every simply-connected finite complex $K$ has finitely generated homotopy groups in this sense.
EDIT 1: I said $\mathcal{F}$ is closed under wedges earlier, but I don't see why now.
EDIT 2: If the answer to original question is "yes", then it is also true rationally. And since the rational question may be easier (I wouldn't be shocked if the experts know the answer to be "no"), I'm explicitly adding it here
QUESTION 2: If $X$ is the rationalization of a simply-connected finite complex, is there a wedge of rational spheres $W$ and a map $W\to X$ which is surjective on $\pi_*$?
FROM A COMMENT BY BEN WIELAND: Question 2 has the following algebraic reformulation, using the Lie model. If we have a differential graded Lie algebra that is finitely generated as a graded Lie algebra, is its homology finitely generated as a graded Lie algebra? (See this question: Is homology finitely generated as an algebra?).
Best Answer
In Donald Kahn's 1966 paper "On stable homotopy modules" (Inventiones 1, pp 375–379, doi:10.1007/BF01389739) he constructs examples of CW complexes with 3 cells, whose stable homotopy groups are not finitely generated as modules over the ring of stable homotopy groups of spheres. I believe this answers in the negative a stable analogue of the question.
Perhaps a Freudenthal suspension theorem argument, together with the stable result, can imply the desired unstable result. I am sorry but I do not have a few minutes to spare to explore this possibility; perhaps others are interested enough to try it out, though. (I am sorry if this incomplete answer is inappropriate for this site, which I am a beginner to. Please feel free to delete this answer if so. I would certainly have made this all a comment rather than an answer, but don't have enough points on this site to leave comments.)