Can an integral domain be embedded in a proper quotient of itself? Does there exist an integral domain $R$ which has a proper ideal $J$ so that there exists an injective ring homomorphism $\phi \colon R \to R/J$?

If yes, what are suitable assumptions on $R$ to exclude such a behaviour? Especially, if $R$ is a finite dimensional complete Noetherian integral domain, can such behaviour occur?
Take the integral domain $R = k[x_1,x_2,\ldots]$ and let $J = (x_1)$. Then $R/J \cong k[x_2,x_3,\ldots]$ which is isomorphic to $R$ because you have infinitely many indeterminates.


  
    Here are some thoughts for your second question: I think if you throw in the assumption that $R$ is a finitely generated $k$-algebra which is also an integral domain, then there is no non-zero prime ideal $\mathfrak{p}$ such that we have an injection $R \to R/\mathfrak{p}$. Indeed, if $\mathfrak{p}$ is non-zero its height is at least one, so that the formula
    $$\operatorname{ht}(\mathfrak{p}) + \dim R/\mathfrak{p} = \dim R$$
    says $\dim R/\mathfrak{p} &lt; \dim R$ and so you can't have an injection $R \to R/\mathfrak{p}$. For finitely generated $k$-algebras $R,S$ that are domains with $R \subseteq S$, we have $\dim R \leq \dim S$ as pointed out by YACP.
EDIT: The following arguments are incomplete (and thus my response might false) - see the comments.

It is enough to assume that $R$ is a finite dimensional Noetherian integral domain in order to exclude this behaviour. In Eisenbud's book (p.219) one can find the following theorem:


  If $R\subset S$ are Noetherian rings such that $S$ is a finitely generated $R$-module, then $\dim R = \dim S$.


So if $R$ is Noetherian, and $\phi$ is as above, then $R/J$ is a finitely generated $\phi(R)$-module, which thus has the same dimension as $R$.

On the other hand, for any ideal $J$ of $R$ one has 

\begin{equation} \dim\  R/J + \operatorname{ht}(J) \le \dim R.\end{equation}

Since any proper ideal has height greater/equal one (here we use the domain assumption), we obtain the contradiction
\begin{equation} \dim R = \dim R/J \le \dim R -1. \end{equation}