No. In the stable homotopy category a retract of a finite cell spectrum is again a finite cell spectrum, but in the weak homotopy category of spaces a retract of a finite cell complex is not necessarily a finite cell complex; there is an obstruction in the kernel of $K_0\mathbb Z[\pi_1(X)]\to K_0\mathbb Z$.
For a simply connected space, finite generation of the direct sum of its integral homology groups implies that it is equivalent to a finite complex. Thus a connected space must become finite after one suspension if its suspension spectrum is finite. The same then follows without assuming connected.
Therefore finiteness of $\Sigma^\infty X$ is equivalent to finiteness of $\Sigma X$, and (as shown by Fernando's answer) this is strictly weaker than finiteness of $X$.