(I wrote this answer in a hurry yesterday, I now fixed a huge number of typos and various mistakes, sorry.)
Well, there are a number of important issues in the question.
First of all, pure quantum states are described by the elements of the complex projective space of a complex Hilbert space $\cal H$. We can therefore work with equivalence classes of unit vectors $[\psi]$, $||\psi||=1$, $\psi \in {\cal H}$. I will denote by ${\cal S}_{\cal H}$ the set of these equivalence classes of unit vectors. A unit vector $\psi\in {\cal H}$ is said state vector, the associated state is the equivalence class $[\psi]:= \{e^{ia}\psi \:|\: a \in \mathbb{R}\}\in {\cal S}_{\cal H}$. A pure state includes all possible physical information on a given quantum system.
If $\langle \cdot|\cdot \rangle$ denotes the inner product in ${\cal H}$,
$p([\psi], [\psi']):=|\langle\psi|\psi' \rangle|^2$ is called transition probability
beteween the state $[\psi]$ and the state $[\psi']$.
Evidently, it is independent of the representative elements chosen in $[\psi]$ and $[\psi']$. So it defines a structure on ${\cal S}_{\cal H}$ (it can be rewrite in terns of a suitable metric on ${\cal S}_{\cal H}$). It is a physical quantity describing a certain type of probability.
The action of $SO(3)$ on pure quantum states is performed in terms of continuous projective representations which preserve the transition probabilities. In other words, it is a map $SO(3) \ni R \mapsto S_R$ where
$S_R: {\cal S}_{\cal H} \to {\cal S}_{\cal H}$ is bijective,
it preserves the $SO(3)$ group structure ($S_I= id$, $S_{R}S_{R'} = S_{RR'}$),
it preserves transition probabilities ($p(S_R[\psi], S_R[\psi'])= p([\psi], [\psi'])$),
it is "continuous"
($SO(3) \ni R \mapsto p([\psi], S_R[\psi'])$ is continuous for every fixed couple of states).
In view of several theorems essentially due to Weyl, Wigner and Bargmann, all these representations can be constructed out of strongly continuous unitary representations $SU(2) \ni g \mapsto U_g \in \mathfrak{B}(\cal H)$ of the universal covering $SU(2)$ of $SO(3)$ acting on ${\cal H}$.
If $\pi : SU(2) \to SO(3)$ is the covering map and $U$ is a representation as above,
$$S_{\pi(g)} [\psi] := [U_g \psi]$$
defines a representation of $SO(3)$ in the space of the states ${\cal S}_{\cal H}$ which respects all requirements above. Varying $U$, one obtains all possible such representations.
Finally, every strongly continuous unitary representation of $SU(2)$ can be decomposed in terms of irreducible representations. These are finite dimensional and of dimension $2j+1$ with $j= 0,1/2, 1, 3/2, \ldots$.
Physically speaking, $j$ is the spin of the considered particle.
If $j$ is semi-integer, then $U_{2\pi}$ is $I$ up to a sign. But it does not matter, since we are interested in $S_{2\pi}=id$ on ${\cal S}_{\cal H}$.
In my view, the claim that rotations of $2\pi$ can be experimentally detected is quite a physical misunderstanding.
The experiments that are claimed to do it work as follows.
One considers a pure state, represented by a vector $\Psi$, which is made of a
superposition of two vectors $\Psi = \psi_1+\psi_2$. The system is a particle with spin $j=1/2$ here.
In the physical space — these vectors are functions $\psi: \mathbb{R}^3 \to \mathbb{C}^2$ — $\psi_1$ and $\psi_2$ have disjoint supports in the physical space, so that in particular $\langle \psi_1|\psi_2\rangle =0$.
Next the system interacts with some physical device which is spatially localized in the support of $\psi_2$ only. If the device is switched on, the net effect is to pass from $\psi_2$ to $-\psi_2$. As the spin is $j=1/2$, this operation corresponds to a "rotation of $2\pi$". Physically speaking, the procedure is in fact performed in terms of the action of a magnetic field on a system with a magnetic momentum. If the system were classical, it would be a rotation of the magnetic moment in the physical space of $2\pi$.
If the device is switched off, the vector $\psi_2$ remains unchanged.
After the operation, the overall vector passes from $\Psi$ to
either $\psi_1 -\psi_2$ or $\psi_1 + \psi_2$. It is evident that the corresponding states are different!
The point is that the device does not act on the whole system. So, it does not rotate of $2\pi$ the whole particle, but just "a part".
What it is true, however is that, if the system were classical we would not observe this change of state.
That is because a classical particle cannot simultaneously stay in two disjoint regions: the supports of $\psi_1$ and $\psi_2$.
A quantum particle (in a sense) can do it, instead!
