How do projective representations act on the QFT vacuum? Let $U:\mathcal{G}\to \mathcal{U}(\mathcal{H})$ be a unitary projective representation of a symmetry group $\mathcal{G}$ on a Hilbert space $\mathcal{H}$. It satisfies the composition rule:
$$U(g_1)U(g_2)=e^{i\phi(g_1,g_2)}U(g_1g_2).\tag{1}$$
Now suppose there is a state $|\Omega\rangle$ which only changes by a phase under the symmetry:
$$\tag{2} U(g)|\Omega\rangle = e^{if(g)}|\Omega\rangle,$$
for some real phases $f(g)$. Then acting on $|\Omega\rangle$ with each side of (1) we find:
$$\phi(g_1,g_2)=-f(g_1g_2)+f(g_1)+f(g_2) \mod 2\pi \tag{3}$$
So $\phi(g_1,g_2)$ is a trivial cocycle. By defining the "improved" symmetry operators $$\tilde{U}(g)=e^{-i f(g)}U(g)\tag{4}$$
we get a true, non-projective representation of $\mathcal{G}$.
To summarise, if a projective representation contains a vector which is left invariant up to phases, then it is not an "intrinsically projective" representation: rescaling the unitaries by phases makes it into a true representation.
But in any QFT without symmetry-breaking there is such an invariant state: the vacuum! So the above argument shows that unbroken symmetries can't be represented projectively in QFT.
My question: What about spinor representations of the Lorentz group - they're projective, aren't they?
 A: *

*Yes, OP is right. More generally, one could consider a projective (not necessarily unitary) representation $\rho:G\to {\rm End }(V)$ where
$$ \rho(g)\rho(h)~=~c(g,h)\rho(gh),\tag{1}$$
and where $c(g,h)\in \mathbb{C}^{\times}$ is a 2-cocycle.
Assume that there exists a 1-dimensional invariant subspace of $V$,
$$\rho(g)|\Omega\rangle~=~b(g)|\Omega\rangle,\qquad 
|\Omega\rangle~\in~V\backslash\{0\}, \qquad 
b(g)~\in~ \mathbb{C}^{\times}~:=~\mathbb{C}\backslash\{0\}, \tag{2}$$
i.e. a 1-dimensional projective subrepresentation. So
$$ c(g,h)~=~b(gh)^{-1}b(g)b(h)\tag{3}$$
is a 2-coboundary, i.e $\rho$ can be lifted to an ordinary linear representation
$$ \tilde{\rho}(g)~:=~b(g)^{-1}\rho(g),\qquad
\tilde{\rho}(g)\tilde{\rho}(h)~=~\tilde{\rho}(gh),\tag{4}$$
such that $\tilde{\rho}$ acts trivially on the vacuum
$$ \left.\tilde{\rho}\right|_{{\rm span}|\Omega\rangle}~=~ \left.{\bf 1}\right|_{{\rm span}|\Omega\rangle}, \tag{5}$$
cf. OP's title question.


*OP's last subquestion about spinor representations for the restricted Lorentz group $SO^+(1,3)$ seems already answered in OP's other post here, i.e. one should consider its double cover $$Spin^+(1,3)~\cong~ SL(2,\mathbb{C})\tag{6}$$ instead.
