Do abelian spinorial prime three manifolds exist? Does there exist a prime 3-manifold such that its mapping class group has an abelian representation in which the 2$\pi$ rotation is represented by -1?
In detail:
Let $M$ be a closed orientable prime 3-manifold.
Let $D_F(M,p)$ be the group of diffeomorphisms of $M$ that fix a point $p$ of $M$ and a frame there. Define the mapping class group (MCG) of $M$ to be the zeroth homotopy group of $D_F(M,p)$. Then the $2\pi$ rotation is an element of MCG that is the equivalence class of the following diffeo, $R_{2\pi}$:
Consider a coordinate ball of radius 2, $B2$, centred on $p$. $R_{2\pi}$ fixes the ball of radius 1, $B_1$, centred on $p$ and everything outside the sphere of radius 2. In between $B_1$ and $B_2$ the $R_{2\pi}$ maps $(x,y,z)\rightarrow(x\cos\theta+y\sin\theta,y\cos\theta−x\sin\theta,z)$ where $\theta$ is a function of $r=\sqrt(x^2+y^2+z^2)$ which increases smoothly and monotonically from 0 to 2$\pi$ as $r$ increases from 1 to 2. The square of the 2$\pi$ rotation is the identity in MCG.
A manifold is spinorial if $\[R_{2\pi}\]$ is non-trivial in MCG.
Background motivation:
This question is interesting because of the 
possibility that fermions can be built on non-trivial 
spatial topology. $M$ is the manifold of a 3-D spatial hypersurface in 
spacetime. The fixed point is the point at infinity (where the metric 
is asymptotically flat) and fixing a frame there has the same effect 
on $\pi_0(D_F)$ as requiring some falloff conditions on the 
diffeomorphisms at infinity or requiring them to be the identity 
outside some ball. The configuration space of General Relativity 
in this asymptotically flat setting is  (space of asymptotically flat metrics on $M$)/$D_F$
and its first homotopy group is isomorphic to (what I called above)
MCG, see http://arxiv.org/abs/math-ph/0606066 (I know it is not the usual definition of MCG). 
The quantum state, on canonical quantisation of General Relativity, carries a unitary 
irreducible representation (UIR) of the MCG and different choices of UIR give different 
physics. Prime 3-manifolds are potentially candidates for 
elementary particles built from pure geometry: topological geons 
(this is speculative!). A prime 3-manifold can 
be the basis for a spinorial particle (i.e. spin 1/2, spin 3/2 
....) if $R_{2\pi}$ is nontrivial. Because particles must be 
able to be pair produced and annihilated, topology change must be 
allowed in the theory which means that the theory should be
quantised in a sum-over-histories framework rather than a canonical 
quantisation framework. Within the sum-over-histories
framework it is challenging to realise nonabelian reps of MCG.
Abelian reps on the other hand are more easily accommodated by attaching phases to 
topologically distinct sectors of the path integral. 
Moreover certain rules that 
would result in a spin-statistics correlation for 
topological geons would also force the reps to be abelian, http://arxiv.org/abs/gr-qc/9609064 (hence the need for abelian reps). However, if there were no spinorial primes with abelian reps 
this would rule out spinorial geons and therefore fermions.
 A: Yes, there exists such a manifold $M$. This follows if there exists aspherical $M$ with $Diff(M)\simeq 0$ (contractible) and $H^2(M;\mathbb{Z}/2\mathbb{Z})=0$. I claim there exists such manifolds. Let's see why such $M$ suffice. 
There are two fibrations:
$$ D_F(M,p) \to Diff(M,p) \to GL(3,\mathbb{R})$$
and
$$ Diff(M,p) \to Diff(M) \to M.$$ 
The first comes from considering the derivative at $p$ of a diffeomorphism fixing $p$. The second comes from considering the image of $p$ under a diffeomorphism of $M$. 
From the long exact sequence of homotopy for a fibration (e.g. 4.41 Hatcher), we have the  exact sequence
$$0=\pi_2(M) \to \pi_1(Diff(M,p)) \to \pi_1(Diff(M))=0 \to \pi_1(M)\to $$
$$\pi_0(Diff(M,p))\to \pi_0(Diff(M))=0.$$
We see that $\pi_0(Diff(M,p))=\pi_1(M)$, and $\pi_1(Diff(M,p))=0$.
Similarly, 
$$0=\pi_1(Diff(M,p))\to \mathbb{Z}/2\mathbb{Z}=\pi_1(GL(3,\mathbb{R})) \to \pi_0(D_F(M,p))\to \pi_0(Diff(M,p))\to 0 .$$
From the second sequence, we see that $\pi_0(D_F(M,p))$ is therefore a $\mathbb{Z}/2\mathbb{Z}$ extension of $\pi_1(M)$. These are classified by $H^2(\pi_1(M);\mathbb{Z}/2\mathbb{Z})=H^2(M;\mathbb{Z}/2\mathbb{Z})=0$ (since any such extension is central). Thus, it is a trivial extension, so $\pi_0(D_F(M,p))=\mathbb{Z}/2\mathbb{Z}\times \pi_1(M)$. The $\mathbb{Z}/2\mathbb{Z}$ factor is generated by $[R_{2\pi}]$ in your notation, so clearly the desired abelian representation exists. 
Now we need to see that a closed orientable  aspherical manifold $M$ with $Diff(M)\simeq 0$ and $H^2(M;\mathbb{Z}/2)=0$ exists. In fact, we may assume $M$ is a hyperbolic homology sphere. For example, take a hyperbolic knot complement with trivial isometry group. Perform Dehn filling of slope $1/k$ for $k$ large to get a closed hyperbolic manifold. This manifold will be an aspherical homology sphere and will have trivial isometry group for $k>>0$ (this follows from Thurston's Dehn surgery theorem and the Margulis lemma, since the isometries must preserve the short core geodesic of the Dehn surgery). By a result of Gabai, $Diff(M)$ will be contractible.  
One can work in somewhat greater generality with hyperbolic 3-manifolds $M$ which have the property that $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})=0$, since the first exact sequence implies that $\pi_0(Diff(M,p))=Aut(\pi_1(M))$. I'm not sure what happens if $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})\neq 0$; I assumed it $=0$ as a convenient way to see that the group $\pi_0(D_F(M,p))$ splits. 
