If I'm not mistaken, there is a counter-example. Have a look at Lazard's second counter-example in:
"Deux mechants contre-exemples" in Séminaire Samuel, Algèbre commutative, 2, 1967-1968.

For any field $k$, Lazard provides a **non-surjective epimorphism** of local $k$-algebras $C\to D$, both of Krull dimension zero, and both of residue field equal to $k$. It is then easy to show that $D$ is also integral over $C$, which is what we need here. Indeed, every $d\in D$ can be written as $d=a+b$ with $a\in k$ and $b$ in the maximal ideal (and unique prime) of $D$, which is therefore nilpotent $b^n=0$, hence trivially integral. Since $a\in k$ is also in $C$, our $d$ is the sum of two integral elements. (Or simply, $D$ is a $k$-algebra, hence a $C$-algebra, generated by nilpotent, hence integral, elements.)

In cash, for those who don't want to click, the rings are constructed as follows:
Consider the local ring in countably many pairs of variables $S=(k[X_i,Y_i]_{i\geq 0})_M$ localized at $M=\langle X_i,Y_i\rangle_{i\geq0}$.
For every $i\geq 0$ choose an integer $p(i) > 2^{i-1}$. Define $J=\langle Y_i-X_{i+1} Y_{i+1}^2 \ ,\ X_i^{p(i)}\rangle_{i\geq0}\subset S$ and define $D=S/J$. Note immediately that $D$ is a local $k$-algebra, say with maximal ideal $m$ and with residue field $D/m\cong S/M\cong k$. Finally, he defines $C$ to be the localization (at $C_0\cap m$) of the subalgebra $C_0:=k[x_i,x_iy_i]_{i\geq 0}\subset D$ where the $x_i$ are the classes of the $X_i$ in $D$ and I let you guess what the $y_i$ are. By construction, the residue field of $C$ is an extension of $k$ which is also a subfield of $D/m=k$, so the residue field of $C$ must be $k$ and we are in the announced situation.

1more comment