Can the complex numbers be realized as a quotient ring?

Yes, $\mathbb{C} = \mathbb{R}[x] / \langle x^2 + 1 \rangle$ is the standard contruction, where $\mathbb{R}[x]$ is the ring of polynomials in one variable with real coefficients; and $\langle x^2 + 1 \rangle$ is the ideal generated by the polynomial $x^2 + 1$.

If you think about it, since $x^2 + 1$ generates the ideal, it's true that

$$x^2 + 1 \equiv 0 \pmod{x^2 + 1}$$

and therefore $x^2 \equiv-1$ so that the equivalence class of the polynomial $x$ is a square root of $-1$.

By mapping $1$ to $1$ and $i$ to $x$ you can get an isomorphism.

[Markup question, Why do I have an extra space in my mod expression above?]

The $\mathbb R[x] / (x^2 + 1)$ solution is what leapt to everyone's mind, but there is an even simpler solution:

$\Bbb C[x]/(x)\cong \Bbb C$.

In both cases, the polynomial whose ideal is being modded out is a maximal ideal of the ring (it would have to be maximal, after all), so it is very much like the integers modulo a prime, as you wished.

Or, for that matter, you could even just say $\Bbb C/\{0\}\cong \Bbb C$, if you don't mind a trivial solution.

One possible definition of $\mathbf C$ is that it is the splitting field of the polynomial $X^2 +1$. As such, it is isomorphic to the quotient ring $\mathbf R[X]/(X^2+1)$ (polynomials with real coefficients, modulo the ideal generated by $X^2+1$). The imaginary number i is then simply the class of X.