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  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 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 = \mathrm{colim} \;R[\alpha\sb 1,\ldots,\alpha\sb n]](http://latex.mathoverflow.net/png?S%20%3D%20%5Cmathrm%7Bcolim%7D%20%5C%3BR%5B%5Calpha%5F1%2C%5Cldots%2C%5Calpha%5Fn%5D)
where the  vary over all the integral elements. By hypothesis each of the modules occurring in the colimit is free so S is flat over R. It now follows that S=R by the following standard argument.
Suppose  is in S, where x,y are in R. Then x is in yS and yS  R = yR by flatness so that y divides x in R also. In particular a is in R.
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. Without R noetherian I am not sure off the top of my head what one can say.