As other answers have pointed out, the answer to the original question is no. However, taking up the idea from David Speyer's comment, both $Gr(3,6)$ and the Schubert divisor $\Delta$ in $Gr(2,7)$ are linear sections of $Gr(3,8)$, by linear subspaces of the same dimension. Specifically, as Schubert varieties inside $Gr(3,8)$, $$Gr(3,6) = \Omega_{(2,2,2)}$$ and $$\Delta = \Omega_{(5,1)}.$$ (I'm using notation where $\Omega_\lambda$ has codimension $|\lambda|$, for $\lambda$ a partition inside the $3 \times (8-3)$ rectangle.) One checks that these two Schubert varieties are both defined by the vanishing of 36 Plücker coordinates (on $Gr(3,8)$).
Taking any curve in the Grassmannian of codimension-36 subspaces inside ${\Bbb P}^N$, where $N = \binom{8}{3}-1$, connecting the two linear spaces cutting out $Gr(3,6)$ and $\Delta$, you should get a flat family having these two as fibers, explaining why they have the same Hilbert polynomial.