If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.
On that component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^B \cdot \times^B P_{\alpha_k} \times^B X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.
(In the familiar case, $X_C = \{ I/I \}$.)