Let $A$ be a normal (i.e. integrally closed in its total ring of quotient) Baer real ring. Suppose furthermore that $T$ is the total integral closure of $A$. If $T$ is actually a finitely generated $A$-module, then $A$ is a real closed $*$ ring and $$A[\sqrt{-1}] = T$$ Let $\p$ be a minimal prime ideal of $A$, then $A/\p$ is a real integral domain. There is a $\tilde\p\in \spec T$ such that $\tilde\p\cap A = \p$ (see [Capco] Remark 109). Then we know, by [HochsterTIC] Theorem 1, that $\qf{T/\tilde\p}$ is an algebraically closed field. Since $T/\tilde\p$ is a finite integral extension of $A/\p$, we can clearly see that $\qf{T/\tilde\p}$ is a finite field extension of $\qf{A/\p}$ (see [Capco] Lemma 46 and [comalg] Proposition 2.1.10). By the classical Artin-Shreier Theorem we then know that $\qf{A/\p}$ is a real closed field.
Since $A$ is normal, for any minimal prime ideal $\p\in \spec A$ , $A/\p$ is integrally closed in its quotient field $\qf{A/\p}$ (which is a real closed field). And so, by [SV] Propositon 2, $A/\p$ is a real closed integral domain for every minimal prime ideal $\p\in \spec A$. By [Capco] Corollary 102 it follows that $A$ is then real closed.
Define $i:=\sqrt{-1_A}$. Then clearly $i\in T$, so by (classical) Artin-Schreier Theorem $\qf{A/\p}$ is a real closed field and $$\qf{A/\p}[i_\p] = \qf{T/\tilde\p}$$ where $i_\p = i\mod\tilde\p$.
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