Yes, a necessary and sufficient condition is that the angle not be a rational multiple of $\pi$.
The problem is equivalent to considering the subgroup generated by $\theta$ in the additive group $\mathbb{R}/\{2k\pi\}$, viewed as the interval $[0,2\pi)$. Under the map $x\mapsto \frac{x}{2\pi}$, we can consider $\mathbb{R}/\mathbb{Z}$, and the subgroup generated by $r+\mathbb{Z}$. The assertion is that the subgroup is dense if and only if $r$ is irrational.
If $r$ is rational, then $\langle r+\mathbb{Z}\rangle$ is finite, so it is not dense.
That the subgroup is dense when $r$ is irrational is a consequence of the Equidistribution theorem. Added. Or more simply, a consequence of Dirichlet's approximation theorem (as noted by Gerry Myerson).
Here is Dirichlet's Theorem:
Theorem. (Dirichlet) For any real number $\alpha$ and positive integer $N$, there exist integers $p$ and $q$, $1\leq q\leq N$, such that
$$\left|q\alpha - p\right|\lt\frac{1}{N+1}.$$
To use it to derive the desired result, let $r$ be an irrational number in $[0,1]$ (corresponding to an angle which is not a rational multiple of $\pi$), and let $\beta\in[0,1]$ be any real number We want to show that for every $\epsilon\gt0$ there exists an integer $n$ such that $nr - \lfloor nr\rfloor$ is within $\epsilon$ of $\beta$ (that is, there is a multiple of the rotation which will land $(0,1)$ within $\epsilon$, along the circle, of $\beta$). Find an integer $M$ such that $\frac{1}{M+1}\lt\epsilon$. By Dirichlet's Theorem, you can find integers $p$ and $q$ such that $|q r - p|\lt \frac{1}{M+1} \lt \epsilon$. That means that if you take $qr$ modulo $1$, you lie within $\frac{1}{M+1}$ of either $0$ or $1$. Either way, by taking multiples of $qr$ you divide the circle into segments of length at most $\frac{1}{M+1}$, and by the Archimedean property some multiple of $qr\bmod 1$ will be strictly larger than $\beta$ (Note that $qr-p$ cannot be equal to $0$, because we are assuming that $r$ is irrational). The distance from $\beta$ to the smallest multiple which is strictly larger is less than $\frac{1}{M+1}$, which shows that there is some multiple of $r\bmod 1$ which is within $\epsilon$ if $\beta$, as desired.
(That is: use Dirichlet's Theorem to get a multiple of the rotation that is smaller in absolute value than $\frac{2\pi}{M+1}$, and note that since $\theta$ is not a rational multiple of $\pi$, this small angle cannot be $0$; then taking multiples of this rotation you get within $\frac{2\pi}{M+1}$ of any point in the circle; pick $M$ sufficiently large so that $\frac{2\pi}{M+1}\lt \epsilon$, and you are done).
So the answer to your question is "yes": the images of $(0,1)$ under successive rotations by an angle $\theta$ are dense in the unit circle if and only if $\theta$ is not a rational multiple of $\pi$.
