To complement Oscar's more systematic answer, let me expand my comment about the case $G = \mathbf{Z}/p\mathbf{Z}$ for a prime number $p$, where the answer is _no_ when $\tilde{K}_0(\mathbf{Z}[G]) \neq 0$.  Candidate non-finite retracts of finite $G$-spectra have already been presented in [the comments][1], to see that they are not equivalent to finite genuine spectra we can use that the natural homomorphisms $$\tilde{K}_0(\mathbf{Z}[G]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$$ 
are both isomorphisms.  The first of these is an isomorphism by a [theorem of Rim][2].  To see that the second map is an isomorphism we use that $\mathbf{Z}[\zeta_p,p^{-1}] = \mathbf{Z}[\zeta_p,(\zeta_p-1)^{-1}]$ and that $(\zeta_p - 1) \subset \mathbf{Z}[\zeta_p]$ is a prime ideal (see e.g., the proofs of Lemma 1.3 and 1.4 on page 2 of [Washington's book][3], in fact $p = (\zeta_p - 1)^{p-1} u$ for $u \in \mathbf{Z}[\zeta_p]^\times$), so exercise 3.8(b) in [chapter I of Weibel's K-book][4] applies.

For any genuine $G$-spectrum $X$, the fixed points $X^e$ for the trivial subgroup come with an action of $G$, so we can form $$C_* (X;\mathbf{Z}[\zeta_p,p^{-1}]) := C_* (X^e;\mathbf{Z}) \otimes_{\mathbf{Z}[G]} \mathbf{Z}[\zeta_p,p^{-1}].$$
This defines an exact functor $C_* (-;\mathbf{Z}[\zeta_p,p^{-1}])$ from the stable $\infty$-category of genuine $G$-spectra to $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$.  This functor sends $\Sigma^\infty_+ (G/e) \mapsto \mathbf{Z}[\zeta_p,p^{-1}]$ and an easy exercise shows that it sends $\Sigma^\infty_+ (G/G) \mapsto 0$.  Therefore a genuine $G$-spectrum which is finite in your sense will be sent to a compact object in $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$ representing $0 \in \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$.

Any element of $\tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}]) $
may be represented by the image under this functor of a retract of 
$ ( \Sigma^\infty_+ G ) ^{\oplus n} $ for some $n$.
  Some such retract then won't be finite when $\tilde{K}_0 (\mathbf{Z}[\zeta_p,p^{-1}]) \neq 0$.


  [1]: https://mathoverflow.net/questions/388784/are-finite-g-spectra-idempotent-complete#comment990097_388784
  [2]: https://doi.org/10.2307/1970033
  [3]: https://doi.org/10.1007/978-1-4612-1934-7
  [4]: https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.I.pdf