Would it be particularly surprising if this were true? If I have understood what you mean correctly (so that since you want to consider integrality in full generality a version of essentialness relative to the image of $R$ is required) I believe I have a proof:

First observe that we can reduce to $R$ local. Indeed one can check normality locally and this preserves minimality of primes (when they survive the localization) and so we can assume $R$ is a local reduced commutative ring with unit.

Now let $P$ be a minimal prime ideal in $R$ and consider the composite $R \rightarrow R/P \rightarrow k(P)$ and suppose that $b$ is an integral element in $k(P)$ over $R/P$ and hence over $R$. As in Jose's comment we know that $b$ is essential over $R/P$ and hence over $R$ (I don't see how to make sense of essential for non-injective morphisms otherwise, maybe I am being dense here). So by hypothesis $R/P[b]$ is a finitely generated projective $R$-module and so is free since $R$ is local. But $\operatorname{Ann}(R/P[b])$ is clearly at least $P$ so $P=0$ (since $R$ is reduced) and $R$ is in fact a domain.

I next claim that in fact $R$ is integrally closed in its field of fractions $K(R)$. To see this lets denote by $S$ the integral closure of $R$ in $K(R)$. Then
$S = \operatorname{colim} R[\alpha_1,\ldots,\alpha_n]$
where the $\alpha_i$ vary over all the integral elements. By hypothesis each of the modules occurring in the colimit is free so $S$ is flat over $R$. In particular, it is flat, finite, and $R \subseteq S$ so that it is faithfully flat over $R$. It now follows that $S=R$ by the following standard argument.

Suppose $a = \frac{x}{y}$ is in $S$, where $x,y$ are in $R$. Then $x$ is in $yS$ and $yS \cap R = yR$ by faithful flatness (we prove this below) so that $y$ divides $x$ in $R$ also. In particular $a$ is in $R$.

Proof that $yS \cap R = yR$: since $S$ is faithfully flat over $R$ we get by changing base a faithfully flat map for any ideal $I$, $R/I \rightarrow S\otimes_R R/I \cong S/IS$
which is injective (since faitfully flat maps are always injective - this follows by using the fact that the kernel of the functor on module categories given by base changing is trivial). In particular we have that $IS \cap R = I$.

In fact I think this gives something stronger. We have shown that the localization at each maximal ideal is a normal domain so in particular $R$ is normal. If $R$ is noetherian it follows that it is a product of finitely many normal domains.