The answer is yes.

The key observation is that rational projective extensions are trivial.

Recall an extension of commutative rings $A \subseteq B$ is called *rational* when for each $b \in B$, the ideal $(A :_A b)$ is dense(=has trivial annihilator) in $A$ and $(A :_A b)b \not= 0$.

**Lemma** Let $A \subseteq B$ a rational extension of rings and $A \subseteq C$ an extension of rings exhibiting $C$ as a projective $A$-module. For any amalgamation $D$ of the span $C \supseteq A \subseteq B$ the multiplication map $B \otimes_A C \rightarrow D$ is injective.

**Proof** Let me just sketch the case that $C$ is a free $A$-module; the generalization to projectives is straightforward. Let $c_i$ be a basis for $C$. If an element $\sum b_i \otimes_A c_i$ maps to $0$, i.e. $\sum b_i c_i = 0$ in $D$, then choose a dense ideal $I$ of $A$ such that $I b_i \subseteq A$. For each element $a \in I$, we get $\sum (a b_i) c_i = 0$. Since $c_i$ is a basis, deduce $a b_i = 0$. Thus $b_i I = 0$, which implies $b_i = 0$.

**Lemma** If $A \subseteq B$ is a rational extension of rings exhibiting $B$ as a projective $A$-module, then $A = B$.

**Proof:** By the previous lemma, $B \otimes_A B \rightarrow B$ is injective, i.e. $B$ is an epimorphism. Note also that $B$ is a faithfully flat $A$-module (it is locally free of nonzero rank). So $A \subseteq B$ is a faithfully flat epimorphism, which is surjective [Stacks Lemma 04VU]

From here there are a number of ways to proceed.

One route is to argue as follows: Suppose that $A$ is reduced and for each ring extension $A \subseteq B$, we have that $A[b]$ is projective whenever $bA[b] \cap A \not= 0$ and $b$ is integral over $A$. In particular this hypothesis will apply to *every* integral element of the extension $A \subseteq Q_{max}(A)$, where $Q_{max}(A)$ denotes the maximal rational extension of $A$, which is a self-injective reduced ring. By the previous lemma, this implies that $A$ is integrally closed in $Q_{max}(A)$. For one thing this implies that $A$ and $Q_{max}(A)$ have the same set of idempotents, so one readily checks that $A$ is Baer because $Q_{max}(A)$ is. *Now we know that $A$ is a Baer ring that is integrally closed in $T(A)$, it's total ring of fractions (which is von Neumann Regular).*

So we could finish with the following.

**Lemma**: Let $A$ be a reduced ring such that $T(R)$ is von Neumann Regular and $A$ is integrally closed in $T(R)$. Then for each minimal prime $P$ of $A$, $A/P$ is integrally closed.

**Proof** Let $\bar{\phi} \in \kappa(P)$ be integral over $A/P$. Lift to a relation $\phi^n + \sum a_i \phi^i = \psi \in PT(A)$. Since $T(A)$ is a VNR, write $\psi = e u$ where $u$ is a unit and $e$ is an idempotent of $T(A)$. Hence $(1 - e) \phi$ is integral over $A$, and by the integrally closed assumption $(1 - e) \phi \in A$. Note also that $e \in A$ by the integrally closed assumption, so $1 - e \equiv 1$ modulo $P$. Hence $\phi \in A/P$.