Extend the cutting plane. In the bottom face, line $DM$ meets line $CB$ extended beyond $B$ at a point $K;$ since $MB \parallel DC$ and $MB = \frac{1}{2}DC,$ segment $MB$ is a midline of triangle $KDC,$ so $B$ is the midpoint of $\overline{CK}$ and $CK = 2.$ The plane also cuts edge $BF$ at a point $P,$ and the piece of the cube cut off past the plane is the pyramid $KDCN$ with the small pyramid $KMBP$ sliced away.

Pyramid $KDCN$ has base $DCN,$ a right triangle with legs $DC = 1$ and $CN = \frac{1}{2},$ and its apex $K$ is at distance $CK = 2$ from the plane of that base, so its volume is $\frac{1}{3} \cdot \frac{1}{4} \cdot 2 = \frac{1}{6}.$ Pyramid $KMBP$ is similar to $KDCN$ with ratio $\frac{KB}{KC} = \frac{1}{2},$ so its volume is $\frac{1}{8} \cdot \frac{1}{6} = \frac{1}{48}.$

The smaller piece therefore has volume $\frac{1}{6} - \frac{1}{48} = \frac{7}{48},$ and the larger piece has volume $1 - \frac{7}{48} = \frac{41}{48},$ giving $p + q = 41 + 48 = 89.$