No nonzero finite spectrum admits an $MU$-module structure. Indeed, suppose $F$ is a finite spectrum with an $MU$-module structure. Then for all $n$, $F$ has a map $v_n:\Sigma^{2p^n-2}F\to F$, which induces an isomorphism on $K(n)_*F$ (there's a subtlety here in that it's not obvious that the $v_n$ map on $F$ and the $v_n$ map on $K(n)$ give rise to the same map on $K(n)\wedge F$; see eg the end of the proof of Lemma 7 of http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf). Thus for each $n$, $F/v_n$ is $K(n)$-acyclic. But $F/v_n$ is finite, so this implies it is also $K(m)$-acyclic for all $m<n$, and so $v_n$ is also an isomorphism on $K(m)_*F$. But by finiteness of $F$, for any $m$ sufficiently large we can find $n>m$ for which $v_n$ must be $0$ on $K(m)_*F$ just for reasons of degree (since by the AHSS for $K(n)_*F$ degenerates). K(m)_*F$). Thus $K(m)_*F=0$ for all sufficiently large $m$, which implies $F=0$.
I would also add that even if you did have a finite spectrum with an $MU$-module structure, it could not possibly be compact as an $MU$-module. Indeed, if it were, after smashing with $H\mathbb{Z}$ it would be a compact $H\mathbb{Z}\wedge MU$-module. But $\pi_*(H\mathbb{Z}\wedge MU)$ is a polynomial ring on infinitely many generators, and so all but finitely many of those generators have to act non-nilpotently on any compact module (basically, any "finite presentation" of a compact module can only involve finitely many of the polynomial generators). Since $\pi_*(H\mathbb{Z}\wedge F)=H_*(F)$ vanishes in all but finitely many degrees for $F$ finite, this is impossible.
