EDIT: The argument in the last sentence is not right. While $R$ is interpretable in $(F,O)$ as the quotient $O/I$, this does not make it definable as a subset of $F$, nor am I sure how to interpret the whole structure $(F,R)$ in $(F,O)$. I’m leaving the answer here in case someone figures how to fix it. (Prehaps there is a version of the AKE principle in a language expanded with a residue field lifting?)
Yes, the theory is decidable.
If $F$ is an ordered field and $R\subseteq F$ a non-cofinal subfield, then $$O=\{x\in F:\exists u\in R\:(-u\le x\le u)\}$$ is a convex valuation ring of $F$, with maximal ideal $$I=\{x\in F:\forall u\in R_{>0}\:(-u\le x\le u)\}.$$ $R$ embeds as a cofinal subfield in the residue field $O/I$; in general, the embedding may be proper, but if $R=\mathbb R$, then $R=O/I$, as $\mathbb R$ is complete.
Thus, let $T$ be the theory of structures $(F,R,+,\cdot,<)$ such that
$F$ is a real-closed field,
$R$ is a non-cofinal subfield of $F$, and
the canonical embedding of $R$ into the residue field $O/I$ as defined above is an isomorphism.
Then $T$ is a recursively axiomatized theory, it is valid in the hyperreal structures $({}^*\mathbb R,\mathbb R)$, and it is complete (see below), hence it is decidable, and axiomatizes the first-order theory of $({}^*\mathbb R,\mathbb R)$.
The completeness of $T$ follows from the Ax–Kochen–Ershov principle, which states that two henselian valued fields of residue characteristic $0$ with elementarily equivalent residue fields and value groups are elementarily equivalent.
First, an easy consequence of basic facts about valued fields is that if $F$ is a real-closed field with a convex valuation ring $O$, then the valued field $(F,O)$ is henselian, the residue field $O/I$ is real-closed, and the value group $F^\times/O^\times$ is divisible.
Then if $(F,R)$ and $(F',R')$ are two models of $T$, their residue fields are elementarily equivalent by completeness of the theory of real-closed fields, and their value groups are elementarily equivalent by completeness of the theory of divisible totally ordered abelian groups, hence the valued fields $(F,O)$ and $(F',O')$ are elementarily equivalent by the AKS principle. Since $R$ is definable in $(F,O)$ by axiom 3, this ensures that $(F,R)$ and $(F',R')$ are elementarily equivalent.