In the classical case of covariant connections, the flatness of a connection means that, locally, one has parallel frames around any point. Now, given a flat contravariant connection $\mathcal{D}$ on a Poisson manifolds $(M,\pi)$, are there local parallel coframes (at least) around regular points of $(M,\pi)$ ?
In general, it is not possible to find parallel frames for a flat contravariant connection, even if the Poisson structure is regular. As an example, let $\pi$ be the zero Poisson structure on a manifold $M$ (which is obviously regular), and let $X$ be a vector field on $M$. Then the operation $$ \nabla f = X \otimes f $$ for $f \in C^\infty(M)$ defines a flat contravariant connection on the trivial bundle $\mathbb{R}\times M \to M$. Indeed, because the Poisson structure is zero, the extra term in the Leibniz rule goes away, so that a contravariant connection must by $C^\infty(M)$linear. Flatness follows from the fact that $X$ is a Poisson vector field; that is, we have the Lie derivative $\mathscr{L}_X\pi = 0$. Nevertheless, there can be no parallel frames in the neighbourhood of any point $p \in M$ at which $X(p) \ne 0$. Indeed, if $\nabla f$ were equal to zero at $p$, we'd have $f(p)X(p) = 0$, so that $f(p) = 0$, but then $f$ cannot be a frame. Recall that in a small enough neighbourhood around any regular point, a Poisson structure is the product of a symplectic manifold and a manifold with the zero Poisson structure. Hence the example above can appear in the neighbourhood of any regular point, provided that the rank of the Poisson structure is less than the dimension of $M$ (i.e. the Poisson structure is not symplectic there). It is worth noting that the existence of parallel frames can be recovered by passing to the symplectic groupoid: suppose that $(M,\pi)$ can be integrated to a sourcesimply connected symplectic groupoid $G$ and let $s,t:G\to M$ be the source and target maps. Then a flat contravariant connection $\nabla$ on the bundle $E$ integrates to a representation of $G$ on $E$. In particular, along each source fibre of $G$, the contravariant connection $\nabla$ gives rise to a flat connection (in the usual sense) on the bundle $t^*(E) \to G$, and so we can find parallel frames along the source fibres. In our example above, in which $\pi = 0$, we have $G = T^*M$, the total space of the cotangent bundle, and the source and target maps are equal to the bundle projection $T^*M \to M$. If we pick a point $p \in M$ the function $\alpha \mapsto \exp(\langle\alpha,X(p)\rangle)$ defined for $\alpha \in T^*_pM = s^{1}(p)$ gives a parallel frame for $t^*(\mathbb{R}\times M) \cong \mathbb{R}\times G$ along $s^{1}(p)$. 

