That is already false when $G$ equals $\mathbb{Z}$. The group ring $\mathbb{Z}G$ is $\mathbb{Z}[t,t^{-1}]$. Let $p$ be a prime integer, and let $I\subset \mathbb{Z}G$ be the ideal $\langle p, t-1 \rangle$. Then $I\otimes_{\mathbb{Z}G}\mathbb{Q}G$ is isomorphic to the principal ideal $\langle t-1 \rangle$, which is free of rank $1$. Yet associated to the short exact sequence, $$0 \to I \to \mathbb{Z}G \to \mathbb{Z}G/I \to 0,$$ consider the long exact sequence of $\text{Ext}^\bullet_{\mathbb{Z}G}(\mathbb{Z}G/I,-)$. The first connecting map quickly gives that $\text{Ext}^1_{\mathbb{Z}G}(\mathbb{Z}G/I,I)$ is nonzero. Therefore $I$ is not projective.

That is already false when $G$ equals $\mathbb{Z}$. The group ring $\mathbb{Z}G$ is $\mathbb{Z}[t,t^{-1}]$. Let $p$ be a prime integer, and let $I\subset \mathbb{Z}G$ be the ideal $\langle p, t-1 \rangle$. Then $I\otimes_{\mathbb{Z}G}\mathbb{Q}G$ is isomorphic to the principal ideal $\langle t-1 \rangle$, which is free of rank $1$. Yet associated to the short exact sequence, $$0 \to I \to \mathbb{Z}G \to \mathbb{Z}G/I \to 0,$$ consider the long exact sequence of $\text{Tor}_\bullet^{\mathbb{Z}G}(\mathbb{Z}G/I,-)$. The second connecting map quickly gives that $$\text{Tor}_1^{\mathbb{Z}G}(\mathbb{Z}G/I,I) = \text{Tor}_2^{\mathbb{Z}G}(\mathbb{Z}G/I,\mathbb{Z}G/I) \cong \mathbb{Z}G/I$$ is nonzero. Therefore $I$ is not projective.

Edit. I corrected this from Ext (incorrect) to Tor (correct).