Yes, I think you can extract Raynaud's result straightforwardly from the Breuil--Kisin theory. Write $k$ for the residue field of $R$. Let's recall how the Breuil--Kisin theory works. Let $\phi : k[[u]] \to k[[u]]$ be the $p$th power map. Then there's an anti-equivalence between the category of $p$-torsion finite flat group schemes over $R$, and the category of finite rank free $k[[u]]$-modules $\mathfrak{M}$ equipped with a $\phi$-semilinear map $\varphi : \mathfrak{M} \to \mathfrak{M}$ such that the image of the ($k[[u]]$-linear) map $\varphi^* : \varphi^* \mathfrak{M} := k[[u]] \otimes_{\phi, k[[u]]} \mathfrak{M} \to \mathfrak{M}$ contains $u^e \mathfrak{M}$. Finally let's recall that the operation of passing to the generic fibre of the group scheme is recovered on the Breuil--Kisin module side by passing from $\mathfrak{M}$ to the étale $\phi$-module $\mathfrak{M}[1/u]$.
So, suppose you have Breuil--Kisin modules $\mathfrak{M} \supset \mathfrak{N}$ with $\mathfrak{M}[1/u] = \mathfrak{N}[1/u]$ (one can always reduce to this case). Choose the least integer $r \ge 0$ such that $\mathfrak{N} \supset u^r \mathfrak{M}$, and take $m \in \mathfrak{M}$ with $n := u^r m \in \mathfrak{N}, u^{r-1} m \not\in \mathfrak{N}$. Since $\varphi(m) \in \mathfrak{M}$ we in particular have $u^r \varphi(m) \in \mathfrak{N}$, and so we require $u^{e+r}\varphi(m)$ to be in the image of $\varphi^*\mathfrak{N}$ under the map $\varphi^*$.
But $u^{e+r}\varphi(m) = \varphi^*(u^{e+r} \otimes_\phi m) = \varphi^*(u^{e-(p-1)r} \otimes_\phi n)$. Since $\varphi^*$ is an isomorphism after inverting $u$, we deduce that $u^{e-(p-1)r} \otimes_\phi n$ must lie in $\varphi^*\mathfrak{N}$, and therefore $e \ge (p-1)r$. If in particular $e < p-1$ then $r=0$ and $\mathfrak{M}=\mathfrak{N}$.
