If $R$ is a domain, it has a fraction field $K$ which in turn has an algebraic closure $\bar K=\Omega$.
This latter field has a well-known Frobenius automorphism $Frob:\Omega \to \Omega: x\mapsto x^p$.
The ring you are looking after is the image of $R$ under its inverse automorphism, namely the ring $R^{1/p}= Frob^{-1}(R)$ You can iterate this process and get rings $R^{1/p},R^{1/p^2},R^{1/p^3} ,\ldots \subset \Omega\;$ whose union is symbolically denoted $R^{1/p^\infty}$ .
If $R$ is not a domain I think you should be very wary and I definitely don't want to say anything about that case.
An example The simplest non trivial example might be the polynomial ring $R=\mathbb F_p[X]$, for which we have $R^{1/p}=\mathbb F_p[X^{1/p}]$.