Let $\bar M_g$ be the DeligneMumford compactifiction of the moduli space of complex genus $g$ curves $M_g$. Is this correct that through every point of the boundary $\bar M_g\setminus M_g$ passes a rational curve that lies in the boundary $\bar M_g\setminus M_g$?
It seems unlikely. Say $g \geq 24$ (or so). There's a divisor $\Delta_g$ on the boundary $M_{2g+1}$ corresponding to curves with a single node consisting of two genus $g$ curves glued at a point. This component is birational to $M_{g,1} \times M_{g,1}$, which is of general type by the assumption on $g$. Thus through a general point of $\Delta_g$ there is no rational curve. For that matter there is no rational curve on $\bar{M}_g$ through this point at all. 


The irreducible components of the normalization $\mathcal{B}^{\nu}$ of the boundary $\mathcal{B}\subset\overline{\mathcal{M}}_{g,n}$ are finite images of the moduli spaces:
Therefore any component of the boundary can be interpred as a moduli space of curves and you question can be rewritten in this way: For which $(g,n)$ is $\overline{\mathcal{M}}_{g,n}$ uniruled? It is well known that $\overline{\mathcal{M}}_{g,n}$ is of general type for $g\geq 24, n\geq 0$. Then it is not uniruled. In lower genus, at the best of my knowledge:


