Answer: no, if the glued scheme is separated, it is affine.

We have a DVR $R$ and an automorphism $\phi:Frac(R)\rightarrow Frac(R)$ such that $\phi(R)$ is not a subset of $R\subset Frac(R)$. We glue $\mathrm{Spec}\:R$ to itself along the generic point by $\phi$. Denote the resulting separated scheme by $X$.

First, let us compute $\Gamma(X, \mathcal{O}_X)$. If we take an open cover, giving a global section of $\mathcal{O}_X$ is the same as giving a section of $\mathcal{O}_X$ on each member of the cover so that the sections agree on the overlaps. So $\Gamma(X, \mathcal{O}_X)$ is identified with the ring $R\cap \phi(R)\subset Frac(R)$.

Note that $R$ is not a subset of $\phi(R)$ because if it were then $\phi^{-1}$ would preserve the order induced by the valuation; then $\phi$ would preserve the order but that is impossible because $\phi(R)$ is not a subset of $R$ by assumption. Thus Theorem 12.2 in "Commutative ring theory" (Matsumura) gives us that $R\cap \phi(R)$ is a PID that has exactly two distinct maximal ideals: $\mathfrak{m}\cap \phi(R)$ and $R\cap \phi(\mathfrak{m})$.

Consider the canonical map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$. A point $x\in X$ is mapped to (the point corresponding to) the kernel of the canonical homomorphism $\Gamma(X, \mathcal{O}_X)\rightarrow k(x)$. So the generic point is mapped to $\{0\}$, and two closed points are mapped $\mathfrak{m}\cap \phi(R)$ and $R\cap \phi(\mathfrak{m})$ respectively. As remarked above, the ideals are distinct so the map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ is injective. Because $X$ and $\mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ both have 3 points, it is bijective and one directly verifies that it is a homeomorphism so $X$ is affine by Stacks 04DE. Admittedly, one could show that the map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ is affine more directly by looking at the inverse images of the affine opens but I like this mode of argumentation more.