Background
Recall that for a real vector bundle, there is a well known integrability theorem.
Theorem. Suppose there is a vector bundle $E$ with fiber $\mathbb R^n$. If $A$ is a flat connection on $E$, then there are local trivializations where the connection matrices of $A$ are identically zero.
Question
Suppose $P$ is a principal $G$-bundle for a Lie group $G$. Then, a connection on $P$ can be described as follows: in each trivialization $(U_\alpha, \psi_\alpha)$ of $P$, there is a 1-form $a_\alpha : U_\alpha \rightarrow Lie(G)$; moreover these 1-forms transform as $a_\beta = \tau a_\alpha \tau^{-1}-d\tau\cdot \tau^{-1}$, where $\tau:U_\alpha\cap U_\beta \rightarrow G$ is the transition map between $\alpha$ and $\beta$.
Statement A. Suppose $(P,a_\alpha)$ as above, such that $a_\alpha$ is a flat connection, i.e., $da_\alpha + a_\alpha \wedge a_\alpha \equiv 0$. Then, given any $U_\alpha$, either $a_\alpha \equiv 0$ or there exists $g:\tilde{U}_\alpha \rightarrow G$ such that $g a_\alpha g^{-1}-dg\cdot g^{-1}\equiv 0$. In words, there is a $G$-gauge in which $a$ is zero.
Question: Is statement A true, at least say for matrix groups $G$? Certainly if $G=GL_n(\mathbb R)$, we know it follows from the Theorem above.
Some thoughts
The proof of the theorem above is essentially solving an ODE for the function $g: I \rightarrow M_n(\mathbb R)$: $$ \frac{dg}{dt} = ga\quad\text{ with }g(0)=Id\,.$$ Notice that if we know $g(t)\in G$ at a given point $t$, then the equation says $g'(t)\in T_gG\subset T_g(M_n\mathbb{R})$ because the connection matrix $a$ always takes values in $Lie(G)$ for a principal $G$-bundle and $g\cdot Lie(G) = T_gG$. So the question is whether $g$ can somehow escape $G$ and go into $M_n(\mathbb R)\setminus G$ even though its tangent vectors were always inside $TG$.