Suppose I have some conserved charge in a 2 dimensional CFT $$Q(|z|)=\int_{w=|z|}\text{d}w\,T(w).\tag{1}$$ The infinitesimal transformation induced on a field $\phi$ at $z$ is then $$[Q(|z|),\phi(z)]=\int_{w=|z|}[T(w),\phi(z)].\tag{2}$$ Books in CFT claim this is not a well defined quantity. On the one hand this seems reasonable to me since at a point in the integral we are taking $[T(z),\phi(z)]$. If $T$ and $\phi$ are distributional one should expect this to run into trouble. On the other hand, in the usual canonical quantization of the scalar field we do not run into such trouble. Indeed, taking $$H(t)=\int\frac{\text{d}^3\vec{p}}{(2\pi)^32E_\vec{p}}\,E_\vec{p}a_\vec{p}^\dagger a_\vec{p},\quad\phi(x)=\int\frac{\text{d}^3\vec{p}}{(2\pi)^32E_\vec{p}}\left(e^{-ipx}a_\vec{p}+e^{ipx}a_\vec{p}^\dagger\right),\tag{3}$$ or $$H(t)=\int\text{d}^3\,\vec{x}\frac{1}{2}\left(\Pi(t,\vec{x})^2+\vec{\nabla}\phi(t,\vec{x})^2+m^2\phi(t,\vec{x})^2\right)\tag{4}$$ one can easily compute $[H(t),\phi(t,\vec{x})]$. Why don't we see singularities in this case?
Another way of phrasing this would be: in the usual canonical quantization of the scalar field in Minkowki spacetime there is a compatibility between the commutators being taken at equal times and the Hamiltonian being constant in time. In CFT one seems to loose this compatibility at some point in the Euclidean field theory. Namely, while the commutators are taken at equal radius, the conservation equation guarantees that the density is holomorphic.
What is the meaning of radially ordering a commutator? Under radial order operators commute.
In eq. (2), I was not referring to the equation you mention in Di Francesco et al. (I think since I couldn't find (6.15b)) The integral on the right hand side contains both $0$ and $w$, which is the integral that corresponds to defining $Q$ as the integral over the density at a fixed time slice.