Yes: just write it out in a local basis: $\sigma_i = \sum f_{ij} E_j$ in some local choice of basis $E_j$, and $\sigma_1(p)=\sigma_2(p)$ means that $f_{1j}(p)=f_{2j}(p)$. Then $d(\sigma_1-\sigma_2)$ is defined at $p$ and represented by $d(f_{1j}-f_{2j}) \otimes E_j$, while $D(\sigma_1-\sigma_2)=\sum d(f_{1j}-f_{2j}) \otimes E_j + (f_{1j}-f_{2j}) \otimes D E_j$ so that at $p$, $D(\sigma_1-\sigma_2)(p)=\sum d(f_{1j}-f_{2j}) \otimes E_j$ represented by the same stuff.

To be careful, $d\sigma_1$ and $d\sigma_2$ are not really defined, even at $p$, because $d\sigma(p)$ is defined for a section $\sigma$ just at the points where $\sigma=0$.