Solution:
Answer: $\mathbf{3}^{\mathbf{2007}}$. If $x$ appears in the sequence, the next term $x^{3} - 3x^{2} + 3$ is the same if and only if $0 = x^{3} - 3x^{2} - x + 3 = (x-3)(x-1)(x+1)$. Moreover, that next term is strictly larger if $x > 3$ and strictly smaller if $x < -1$. It follows that no values of $a_{0}$ with $|a_{0} - 1| > 2$ yield $a_{0} = a_{2007}$.

Now suppose $a_{0} = a_{2007}$ and write $a_{0} = 1 + e^{\alpha i} + e^{-\alpha i}$; the values $a_{0}$ we seek will be in bijective correspondence with solutions $\alpha$ where $0 \leq \alpha \leq \pi$. Then
$$a_{1} = (a_{0} - 1)^{3} - 3a_{0} + 4 = e^{3\alpha i} + 3e^{\alpha i} + 3e^{-\alpha i} + e^{-3\alpha i} - 3e^{\alpha i} - 3e^{-\alpha i} - 3 + 4 = e^{3\alpha i} + e^{-3\alpha i} + 1$$
and an easy inductive argument gives $a_{2007} = e^{3^{2007} \alpha i} + e^{-3^{2007} \alpha i} + 1$. It follows that $a_{0} = a_{2007}$ is equivalent to $\cos(\alpha) = \cos(3^{2007} \alpha)$. Now,
$$\cos(3^{2007} \alpha) - \cos(\alpha) = 2 \sin\left(\frac{3^{2007} + 1}{2} \alpha\right) \sin\left(\frac{3^{2007} - 1}{2} \alpha\right)$$
so since $\sin(kx) = 0$ for a positive integer $k$ if and only if $x$ is a multiple of $\frac{\pi}{k}$, the solutions $\alpha$ are $\{0, \frac{2\pi}{3^{2007} - 1}, \frac{4\pi}{3^{2007} - 1}, \ldots, \pi\} \cup \{0, \frac{2\pi}{3^{2007} + 1}, \ldots, \pi\}$. Because our values $k$ are consecutive, these sets overlap only at $0$ and $\pi$, so there are $3^{2007}$ distinct $\alpha$.