In case you'd like an example which is geometrically integral (as Karl's isn't), here's one. Let $A$ be the subring of $\mathbb{R}[x]$ consisting of polynomials which take real values at $\pm i$. Explicitly, $A$ is generated by $u:=x^2+1$ and $v:=x(x^2+1)$, and can be written as $\mathbb{R}[u,v]/\langle v^2+u=u^2 v^2+u^2=u^3 \rangle$.
As you would expect, the ideal of polynomials vanishing at $\pm i$ (also known as the ideal $\langle u,v \rangle$) has residue field $\mathbb{R}$. The normalization is $\mathbb{R}[x]$ (in terms of $u$ and $v$, we have $x = v/u$), the preimage of $\langle u,v \rangle$ is $\langle x^2+1 \rangle$ and the corresponding residue field is $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.
In case you'd like an example which is geometrically integral (as Karl's isn't), here's one. Let $A$ be the subring of $\mathbb{R}[x]$ consisting of polynomials which take real values at $\pm i$. Explicitly, $A$ is generated by $u:=x^2+1$ and $v:=x(x^2+1)$, and can be written as $\mathbb{R}[u,v]/\langle v^2+u=u^2 \rangle$.
As you would expect, the ideal of polynomials vanishing at $\pm i$ (also known as the ideal $\langle u,v \rangle$) has residue field $\mathbb{R}$. The normalization is $\mathbb{R}[x]$ (in terms of $u$ and $v$, we have $x = v/u$), the preimage of $\langle u,v \rangle$ is $\langle x^2+1 \rangle$ and the corresponding residue field is $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.