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, 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 is a theorem of Rim, the second is because $p\mathbf{Z}[\zeta_p] \subset \mathbf{Z}[\zeta_p]$ is a prime ideal (see e.g., exercise 3.8(b) in chapter I of Weibel's K-book).
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 $\tilde{K}_0 (\mathbf{Z}[\zeta_p,p^{-1}]) \neq 0$.