QM rotation operator I have seen the proof that for fermions a rotation of $2 \pi$ does not return a spin angular momentum eigenstate to its original form, but instead multiplies the wavefunction by $-1$.
Here is an abbreviated derivation:
\begin{equation}
U(\vec \alpha) = \exp (-i \vec{\alpha} \cdot \vec{J}) = \sum_{n=0}^{\infty} \frac{(-i)^n \left[\vec{\alpha} \cdot \frac{1}{2} \vec{\sigma}^n \right]}{n!} = I \cos{\left(\frac{\alpha}{2} \right)} - i \hat{\alpha} \cdot \vec{\sigma} \sin{\left(\frac{\alpha}{2} \right)}
\end{equation}
where I have used the fact that $\vec{J} = \vec{S} = \frac{1}{2} \vec \sigma$ for a spin-half particle at rest. For $|\alpha| = 2 \pi$, we have:
\begin{equation}
U(\vec \alpha) = -I
\end{equation}
I understand this is a special state for fermionic spin, and not the case for a state with zero spin.
Is there an equivalent proof that a rotation of $2 \pi$ will  return a state with just orbital angular momentum to its original state?
 A: The identity you used,
$$
\exp(i\theta \, \hat s)=\cos(\theta)+i\sin(\theta)\,\hat s,
\tag{$\ast$}
$$
is crucially dependent on the operator $\hat s$ being idempotent, and particularly on the fact that $\hat{s}^2=\mathbb1$. This is generally not the case for angular momenta other than spin-1/2. In general, the total angular momentum is a scalar within the representation (i.e. $\hat{\mathbf J}^2=\hbar^2j(j+1)\times\mathbb 1$) but outside of spin-1/2 it is no longer the case that each component is idempotent, so if $\hat{\mathbf n}$ is a unit vector, $\hat{\mathbf n}\cdot\hat{\mathbf J}$ no longer squares to a multiple of $\mathbb 1$. In that case, the rotation operator is still given by
$$
\hat U(\vec \alpha)=\exp(-i\vec \alpha\cdot\hat{\mathbf J}/\hbar)
$$
but no further simplifications are possible.

That said, if you have a specific $j$ you want to investigate, there is probably an analogous formula to $(\ast)$, because the $(2j+1)$th and higher powers must be linear combinations of the first $2j$ powers and the identity, so you can again fold the exponential series into $2j+1$ terms,
$$
\exp(-i\theta \hat{\mathbf n}\cdot\hat{\mathbf J})=\sum_{k=0}^{2j}f_k(\theta) (\hat{\mathbf n}\cdot\hat{\mathbf J})^k,
$$
where the $f_k$ are given by appropriate series. Depending on what comes out, this may or may not be useful, but again it is a lot of work for each specific $j$ you're interested in.
To be a bit more explicit, let me show how this works for the simplest nontrivial orbital angular momentum, $j=1$. If you work in the $z$ direction you get, for the different powers of $\hat J_z$,
$$
\mathbb 1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}
,\quad
\hat J_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
,\quad
\hat J_z^2=\hbar^2\begin{pmatrix}1&0&0\\0&0&0\\0&0&1\end{pmatrix}
,\quad
$$
and it's only then that you get stuck in a loop: $\hat J_z^{2+n}=\hbar^{n+1} \hat J_z$ for odd $n\geq1$ and $\hat J_z^{2+n}=\hbar^{n} \hat J_z^2$ for even $n\geq0$. This means that the exponential folds again, but not as neatly:
$$
\exp(-i\theta \hat J_z/\hbar)
=\sum_{n=0}^\infty\frac{(-i\theta \hat J_z/\hbar)^n}{n!}
=
\mathbb 1
-\frac{i\sin(\theta)}{\hbar} \hat J_z
+\frac{\cos(\theta)-1}{\hbar^2}\hat J_z^2.
$$
This is similar, but not quite the same, as the original $(\ast)$. It does yield the desired invariance after $2\pi$ rotations, but it required a lot of work for a single $j$.

This does leave you with the need to prove that for general, integral $j$ all states will return to themselves after a $2\pi$ rotation. The proof is somewhat different, though, and it relies on the fact that the eigenvalues of $\mathbf J$ are all integers. In particular, for any given unit vector $\hat{\mathbf n}$ there will be a basis $\{|m⟩=|j,m,\hat{\mathbf n}⟩\}_m$ of eigenstates of $\hat{\mathbf n}\cdot\hat{\mathbf J}$ with integer eigenvalues:
$$
\hat{\mathbf n}\cdot\hat{\mathbf J}|m⟩=\hbar m|m⟩,\quad m=-j,\ldots,j.
$$
This lets you calculate the action of $\hat U(\vec \alpha)$ on each eigenstate:
$$
\hat U(\vec \alpha)|m⟩
=\exp(-i\vec \alpha\cdot\hat{\mathbf J}/\hbar)|m⟩
=\exp(-i\alpha\hat{\mathbf n}\cdot\hat{\mathbf J}/\hbar)|m⟩
=\exp(-im\alpha)|m⟩.
$$
For $\alpha=2\pi$ and integer $m$, this is exactly $|m⟩$, with no added phase. Since any arbitrary state $|\psi⟩$ can be expressed as a linear combination of the $|m⟩$ and those are unchanged by $\hat U(\vec \alpha)$, it follows that $|\psi⟩$ itself is also unchanged by $\hat U(\vec \alpha)$.
Finally, you should note that this proof also works to show that half-integral $j$s produce $\pi$ phases upon $2\pi$ rotations, since then every $m$ is half-integral, and $e^{-i2\pi m}\equiv -1$ regardless of $m$. Your original proof, though, also fails for $j\geq 3/2$, since the idempotency condition is also not fulfilled.
A: The point is that the spin operator is defined to be (1/2) times SU(2) generator while the orbital angular momentum is defined to be only SU(2)(or SO(3), is the same) generator. The proof is the same, and is " representation independent", in the sense that the structure of identity multiplied by something plus a linear combination of sigma matrices multiplied by something else is the same in any representation and depend only on the algebra of SU(2) (or SO(3)). The "something" structure depend by the definition of operator with which you are dealing. So if there is no (1/2) behind SU(2) generator there is no $\alpha/2$ in the trigonometric structure.
A: You can use similar derivation, but with momentum operator J for integer spin instead of Pauli matrix $1/2\sigma$. Then you get integer coefficient in from of $\alpha$ instead of half-integer in the final formula... and the same state after $2\pi$ rotation.
