Let $R$ be an $\mathbf{F}_p$-algebra. Kunz's theorem says that if $R$ is Noetherian, then the Frobenius of $R$ is flat iff $R$ is regular. Following the philosophy that valuation rings often behave like regular Noetherian rings, it is shown in [Datta and Smith - Valuations and Frobenius](https://arxiv.org/abs/1507.06009) that if $R$ is a valuation ring, then the Frobenius of $R$ is flat.

I'm wondering about a mixed characteristic variant of this. Let $R$ be a $p$-complete valuation ring (in which $p$ is nonzero and not invertible) with maximal ideal $\mathfrak{m}$, and let $\pi \in \mathfrak{m}$ be a regular element such that $\pi^p$ divides $p$. Is the Frobenius map $R/\pi \to R/\pi^p$ flat?