Let $M$ be a compact Reimannian manifold and let $(E,h)$ be a Hermitian vector bundle over $M$. Let $A^k(M,E)$ denote the space of $E$-valued $k$-forms, i.e. smooth section of the bundle $\bigwedge^kT^*M\otimes E$.
Let $D$ be a flat Hermitian connection on $(E,h)$ (where 'Hermitian' means $D$ is compatible with the metric $h$: $dh(s,t)=h(Ds,t)+h(s,Dt)$, $\forall s,t\in A^0(M,E)$), and let $D^*$ denote its formal adjoint (i.e. $(D\alpha,\beta)=(\alpha,D^*\beta)$ where $(-,-)$ is the $L^2$-inner product of $A^*(M,E)=\bigoplus_kA^k(M,E)$ induced by $h$). Then one can define the Laplacian w.r.t. to $D$: $\Delta_D=-D^*D-DD^*$.
A Hodge decomposition theorem on $A^*(M,E)$: $$A^*(M,E)=ker\Delta_D\oplus im D\oplus im D^*$$ is demonstrated in Demailly's $L^2$ Hodge Theory and Vanishing Theorems (the first chapter of the book 'Introduction to Hodge Theory'), Section 4. The proof is actually the same as the proof in the case without $E$: to show the formally self-adjoint operator $\Delta_D$ is elliptic and to use linear elliptic PDE theory.
Now I wonder wether the condition that $D$ is Hermitian can be removed. It seems to me that the only place where this condition is used is to calculate $D^*=(-1)^{...}\star D \star$. This formula implies the principal symbol $\sigma_{D^*}(\xi)=-\xi_{\sharp}\lrcorner$, then it follows that $\sigma_{\Delta_D}(\xi)=|\xi|^2$: the differential operator $\Delta_D$ is elliptic. But using antilinear Hodge operators $\bar\star:A^{p,q}(M,E)\leftrightarrow A^{n-p,n-q}(M,E^*):\bar\star$, we have $D^*=(-1)^{...}\bar\star D_{E^*}\bar\star$ even $D$ is not Hermitian. Then the same consequences still follow.