In a certain sense, what you said that every operator might be represented in position representation as a integral operator may be true for many operators only if you allow distributions to be used, and even though that's not always the case.
You are kind of confusing things when you say about the dual of distributions. What is a distribution is the representation of $|r\rangle$, which is not a vector of the Hilbert space $\mathcal{H}$, that is $\langle x |r\rangle=\delta(r-x)$. The problem about using this notation is that is that this lacks a lot of rigor and sometimes if you do not pay attention you may get into trouble.
What may be hard in general is to find the position representation of an operator. For example, take the free propagator $U(t)=e^{-itH}$, with $H=-\Delta$ being the free hamiltonian. It's possible to show that
$$U\psi(x)=(e^{-itH}\psi)(x)=\frac{1}{\sqrt {4\pi it}}\int_\Bbb{R} e^{i\frac{(x-y)^2}{4t}}\psi(y)dy,$$
which shows that $U(t)$ may be represented as an integral operator in position representation with kernel $K(x)=\dfrac{1}{\sqrt {4\pi it}}e^{i\frac{x^2}{4t}}$. One way to find the representation of $F$ is to change the basis where the operator is a multiplication operator. This is always possible if $F=f(T)$, where $f$ is a measurable function and $T$ is a self-adjoint operator. In that case, the spectral theorem states that
$$F=\int_\Bbb{R} f(t)dP^T(t),$$
where $P^T$ is the resolution of the identity of $T$. In practice it's possible to write $F=U^{-1}fU$, where $U$ is the unitary map that maps the position representation to the basis where $T$ is a diagonal operator. For example, if $T=P$, the momentum operator, then $U=\mathcal{F}$ is the Fourier transform. So if you want to find the position representation of a function $F=f(P)$ you may proceed like:
$$(F\psi)(x)=(\mathcal{F}^{-1}f(p)\mathcal{F}\psi)(x)=\frac{1}{2\pi}\int e^{ipx}\left(f(p)\int e^{-ipy}\psi(y)dy\right)dp=\int K(x,y)\psi(y)dy,$$
with
$$K(x,y)=\frac{1}{2\pi}\int e^{-ip(x-y)}f(p)dp$$
The last step is formal, because for us to change the order of integration, the integrals must exists, and that's not always the case, and you might get distributions, like is the case if $f(p)\in \mathcal{S}'(\Bbb{R})$, that is $f$ is a tempered distribution. E.g, taking $F=P$, that is $f(p)=p$, we find $K(x,y)=i\delta'(x-y)$. But if you take for example $f(p)=e^p$, the kernel $K(x,y)$ is not defined even in the sense of tempered distributions, although the operator $F=\exp(P)$ is well defined for a dense domain.