Ionizing radiation in thermal radiation According to the black-body radiation equation, the spectrum extends to infinitely high frequency (although its intensity gets small quickly towards high frequency).
(1) How do you roughly estimate the ionizing radiation power in common high power thermal sources like a 2KW heater without fan, then make sure it is safe considering it is used for years instead of 0.1 second/year in the case of DR X-ray scan? (First the portion of power getting radiated instead of convected needs to be estimated, second this seems not to be black-body and will it have a similar radiation curve that extends to infinite frequency?)
(2) In the photoelectric effect, there are things like cut-off frequency; so, why doesn't thermal radiation - which is quantum mechanical in microscopic level - have a cut-off frequency, i.e. it doesn't radiate X-ray and gamma-ray at all?
 A: (1) The element in most heaters appears to be a cooler color/temperature than the sun - and only similarly warm.  So, it is safer than the sun.
(2) The photo-electric effect reveals that below a cut-off frequency there is not enough energy to release an electron, so the energy eventually ends up as heat.  Above the cut-off an electron may be freed.  But all frequencies are involved and can do something.
A: If you plug in the values of $T=1000$K, $\nu=10^{15}$Hz into the spectral radiance formula of a blackbody (Plank’s law):
$$B_{\nu}(\nu, T)
    =
    \cfrac{2h \nu^3}{c^2}
    \frac{1}{\exp\left(\frac{h \nu}{k_BT}\right) - 1}$$
where $B_{\nu}(\nu, T)$ is the power delivered per unit area per solid angle at a frequency range centred at $\nu$ by a blackbody at temperature $T$ and do the following integral
$$P\left(T,\nu_0\right)=\int_{\nu_0}^{\infty} \cfrac{2h \nu^3}{c^2}
    \frac{d\nu}{\exp\left(\frac{h \nu}{k_BT}\right) - 1}$$
Here since the argument of the exponential is $\sim 48$, we can approximate the integral to
$$P\left(T,\nu_0\right)=\int_{\nu_0}^{\infty} \cfrac{2h \nu^3}{c^2}\exp\left(-\frac{h \nu}{k_BT}\right)d\nu$$
This is a straightforward integral which can be done by parts to get the total ionising radiation power to be 
$$P\left(1000\text K,10^{15}\text{Hz}\right)=1.69\times10^{-16}\frac{\text{W}}{\text{m}^2\cdot\text{st}}$$
This as you can see, is minuscule. 

In photoelectric effect, we are knocking off an electron from the bulk using the energy of photons. The excess energy is converted into kinetic energy of the electron. Now consider the reverse. High energy electron is getting captured by the bulk. So now a photon is released depending on how much energy the electron lost. Since the electron before capture can have arbitrary energy, the emitted photon can have arbitrary energy. Thus there is no cutoff for radiation. 
The cutoff comes in the case of photoelectric effect due to the restriction on allowed energies of the bound electron. However there is no such restriction on free electron.  
A: Planck's law says:
$$B_\nu(\nu, T) = \frac{2h\nu^3}{c^2} \frac1{\exp\left(\frac{h\nu}{kT}\right) - 1}.$$
This is the power emitted by a black body of temperature $T$ at frequency $\nu$. If we consider a black object (thus highest possible purely thermal emission) of temperature $700°\mathrm C$ (an estimate of temperature of a heater), it'll emit total power of
$$P_\text{total}(700°\mathrm C)=\pi \int\limits_0^\infty \mathrm{d}\nu B_\nu(\nu,700°\mathrm C)=13.6\,\frac{\mathrm{kW}}{\mathrm{m}^2}.$$
(This can also be calculated via the Stefan-Boltzmann law.)
If we now want to find the power of only ionizing radiation, we should instead integrate in the range of $\nu\in[\nu_0,\infty)$, where $\nu_0$ could be taken somewhere in the UV range, e.g. lower border of UVB, i.e. $950\,\mathrm{THz}$. Then we have:
$$P_\text{hard}(700°\mathrm C)=\pi \int\limits_{950\,\mathrm{THz}}^\infty \mathrm{d}\nu B_\nu(\nu,700°\mathrm C)=3.1\times 10^{-20}\,\frac{\mathrm{W}}{\mathrm{m}^2}.$$
So a surface of $1\,\mathrm{m}^2$ will emit $~10^{-20}\,\mathrm{W}$. This is totally negligible, even in the course of a hundred years.
A: for 1) In this link there is the black body formula of Planck's law that gives the 

the power per unit solid angle and per unit of area normal to the propagation

So for a given temperature of the body one can calculate the power, in your case for high enough frequencies, take a $Δ(ν)$  . 
Once you have the power you will have the energy per second. Divide the power with  the average $hν$ of your  $Δ(ν)$ and you can estimate how many photons of that approximate frequency radiate per second. 
You will find that one has to wait for a loooo..ng time for an X-ray photon to come our of your radiator. Better worry about the Xrays created by the muons passing your body at the rate of 1 every $cm^2$ every minute, much worse when you fly in an airplane.
for 2) Quantum mechanics gives probabilities for limits. When the probability gets very very small, that is a limit.To get enough energy for an x-ray the material would have to synchronize so that random vibrations would quantum mechanically allow for a quantum level to exist to have enough energy for an x-ray photon to come out.How improbable that is is reflected in the  Planck formula.
Also a limit is given by energy conservation laws , also included in the formula.
The basic line is that it is very improbable, as the other answers state.
A: 
How to roughly estimate the ionizing radiation power in common high
  power thermal source like 2KW heater without fan then make sure it is
  safe considering it is used for years instead of 0.1 second/year in
  case of DR Xray scan?

A common 2 kW resistance heater does not emit ionizing radiation. It's radiant heat is primarily infra red. The range for infrared  radiation is about 430 x 10$^{12}$Hz to 300 x 10$^9$ Hz. It is considered non-ionizing radiation.
Ionizing radiation begins around 3 x 10$^{15}$ Hz. X-radiation is in the range of 3 x 10$^{16}$ to 2 x 10$^{19}$ Hz.  
Bottom line: Infrared heaters do not pose the health risks associated with ionizing radiation like X-radiation.
Hope this helps.
