Formal models of rigid discs of any radii sorry if this is a too vague. For $K$ some non-archimedean discretely valued field the rigid disc of radius 1 $\mathrm{Sp} \, K\langle T \rangle$ has a formal model $\mathrm{Spf} \, K^{\circ} \{T\}$. I just read a bit about Berkovich spaces and his theory allows discs of any radius $r$: $K\langle r^{-1} T \rangle$. Is there some theory of "formal models" in this context?
 A: Not really an answer as I am not familiar with Berkovich stuff.
But, ... in classical rigid geometry, one also has a disc of any rational radius $r$. If $r=\frac{a}{b}$ and $\pi$ is a uniformizer of $\mathcal{O}_K=K^{\circ}$, then $\textrm{Spf}(K^{\circ}\{T, U\}/(T^b-\pi^{a}U))$ is a formal model (Raynaud style) of the subdisc $K\langle r^{-1}T\rangle$. However, it may not be as nice as you may want. For example, its special fiber is not reduced. It becomes so after a sufficiently large extension of $K$.
Since the formal models are basically the same in both theories, the discs in Berkovich with radii in $|K|$ should have these kinds of models.
A: The formal models used in Berkovich geometry are the same that are used in rigid geometry. In particular, their generic fibers are built from spectra of quotients of Tate algebras of the form $K\langle T_1,\dotsc,T_n\rangle$, called strictly $K$-affinoid in Berkovich's terminology.
To answer your question, one checks that, for $r\in\mathbf{R}_{>0}$, the algebra $K\langle r^{-1}T\rangle$ is strictly $K$-affinoid if, and only if, $r\in \lvert K^*\rvert^\mathbf{Q}$. (For the if direction, use the trick in A.B.'s answer. For the only if direction, use that the supremum of a function on a strictly $K$-affinoid space belongs to $\lvert K^*\rvert^\mathbf{Q}$.) So, you will not get any formal model for $K\langle r^{-1}T\rangle$ if $r\notin \lvert K^*\rvert^\mathbf{Q}$.
On the other hand, as pointed out by Piotr Achinger in his comments, Michael Temkin worked out a theory of reduction for all $K$-affinoids, strict or not. It has properties very similar to the usual reduction of affinoids, with objects only slightly more complicated, actually graded versions of the usual objects. For $r\notin \lvert K^*\rvert^\mathbf{Q}$, the reduction of $K\langle r^{-1}T\rangle$ is $\tilde{K}[r^{-1}\tilde{T}]$, where the notation means that $\tilde{T}$ has degree (with respect to the graduation) $r$. It follows that $\tilde{K}[r^{-1}\tilde{T}]$ has few homogeneous elements, only the monomials. As a result, its graded spectrum, which is Temkin's reduction of $K\langle r^{-1}T\rangle$, has only two points: the generic point $(0)$ and the closed point $(\tilde T)$.
