A single-qubit rotation looks like $U(\theta,\mathbf{n})=\exp(-i \theta\mathbf{n}\cdot\pmb{\sigma})$ for the vector of Pauli matrices $\pmb{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$. Here we assume $\mathbf{n}$ is normalized to unity; otherwise we can equivalently define some vector $\mathbf{r}=\theta\mathbf{n}$ and say $U(\mathbf{r})=\exp(-i \mathbf{r}\cdot\pmb{\sigma})$.
It indeed holds that every unitary can be written as the exponential of some Hermitian "generator"
$$U(G)=\exp(-iG).$$ There is always a generalized basis in which this generator can be written for any number of qubits (see Wikipedia), with $G=\mathbf{r}\cdot\pmb{\Sigma}$, where I'm using $$\pmb{\Sigma}=(\sigma_0\otimes\sigma_0\otimes\cdots,\sigma_0\otimes\sigma_1\otimes\cdots,\cdots,\sigma_3\otimes\sigma_3\otimes\cdots)$$ to represent all possible combinations of Pauli matrices. If we normalize $\mathbf{r}$, we get to write
$$U(\theta,\mathbf{n})=\exp(-i\theta \mathbf{n}\cdot\pmb{\Sigma})$$ in any dimension, so long as we use all of the basis generators in $\pmb{\Sigma}$ and we make $\mathbf{n}$ have sufficient dimensions ($4^N$ for $N$ qubits, minus one for the global phase).
Can we interpret $\theta$ as a rotation angle in the sense that the same unitary is enacted for $\theta+2\pi$? All unitaries have eigenvalues with modulus one, of the form $e^{i\lambda}$, so we know that adding phases will do something sensible. Start by finding an eigenbasis of the Hermitian operator $\mathbf{n}\cdot\boldsymbol{\Sigma}$:
$$\mathbf{n}\cdot\boldsymbol{\Sigma}|\psi_k\rangle=\lambda_k|\psi_k\rangle,$$ where each eigenvalue $\lambda_k$ and eigenvector $|\psi_k\rangle$ may depend on $\mathbf{n}$ and $\boldsymbol{\Sigma}$. We can then write the spectral decomposition of $U$ in this eigenbasis as
$$U(\theta,\mathbf{n})=\sum_{k}e^{-i\theta \lambda_k}|\psi_k\rangle\langle\psi_k|.$$
Let's stare at this for a few moments. If $\theta\to\theta +2\pi$, does $U$ stay the same? Only if $\exp(-2\pi i\lambda_k)=1$ for all $\lambda_k$! Maybe we'll permit $U$ to change by a global phase and still enact the same transformation, so we need the slightly more generous but still restrictive condition $\exp(-2\pi i\lambda_k)=e^{i\varphi}$ for all $\lambda_k$ and some real phase $\varphi$. Not so easy for a general distribution of eigenvalues.
If we keep staring, we realize that each $e^{-i\theta\lambda_k}$ is rotating around in the complex plane with period $2\pi/\lambda_k$, so we are seeking another rotation angle $\theta+\phi$ that matches up all of those rotations to each return to their starting point. If all of the ratios $\lambda_k/\lambda_l$ are rational numbers [or even more technically if this is true for any $(\lambda_k+E)/(\lambda_l+E)$ all for the same $E$ because the absolute "eigenenergies" of the generator just lead to global phases], there will always be some angle $\theta+\phi$ for which the phases all align and the unitary returns to the same value as $\theta$ (mathematically, this is still just looking for $\exp(-i\phi\lambda_k)=1$ for all $\lambda_k$). In general, the ratios may be irrational, so the unitary may never return to the same point, but in many special cases it does.
What kind of special cases are we talking about? Well if you've heard of angular momentum operators, you know the eigenenergies are spaced by half integers, so you know the rotations will eventually return to the same starting point. If all you tell me is that you have qubits, I can concoct an arbitrary unitary that either does or does not return to its starting point:
$$U_{\mathrm{return}}=e^{-i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N}+e^{-2i\theta}\left(|1\rangle\langle 1|\right)^{\otimes N}+e^{-3i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N/2}\otimes \left(|1\rangle\langle 1|\right)^{\otimes N/2}$$ returns to its starting point when $\theta\to \theta+2\pi$ and
$$U_{\mathrm{no\,return}}=e^{-i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N}+e^{-\sqrt{2}i\theta}\left(|1\rangle\langle 1|\right)^{\otimes N}+e^{-\sqrt{3}i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N/2}\otimes \left(|1\rangle\langle 1|\right)^{\otimes N/2}$$ never returns to its starting point for any $\theta\to\theta^\prime$.
