Can hot food ever emit x-rays or gamma rays?
In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen).
Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The energy of one photon of 30 petahertz radiation is:
$$E=h\nu = \left(6.626\times 10^{-34}\mathrm{\ J\cdot s}\right)\left(30\times 10^{16}\ \mathrm{s^{-1}}\right) = 1.988 \times 10^{-16}\ \mathrm{J}$$
This is not a lot of energy! However, a single photon is boring. Let's consider a mole of photons. This will also ease comparison with other phenomena, whose energies are listed per mole of events.
$$1.988 \times 10^{-16}\ \mathrm{J} \times 6.022\times 10^{23}\ \mathrm{mol^{-1}}=1.197\times 10^8\ \mathrm{J\cdot mol^{-1}}$$
In theory, if you could pump that much energy into something, you should get some high energy photons out. In practice, it does not work that way. Other stuff happens first. To simplify our example, let's just consider 1 mole of water (18.0 grams) and heat it up. The fate of basically any other matter will be the same, but the energy required will vary a bit.
First, adding energy heats the water. If we start at room temperature $\left(20\ ^\circ\mathrm{C}\right)$, it takes $80\ ^\circ\mathrm{C}\times 18\ \mathrm{g}\times 4.184\ \mathrm{J\cdot g^{-1}\cdot ^\circ C^{-1}}=6025\ \mathrm{J}$ to heat that water to boiling. It takes 40.66 kJ to convert the water into gas. Neither of these puts a big dent in our energy. It takes further energy to heat the water vapor again, but let's see how far we need to take it.
Once we get enough energy into our sample of water, the molecules start to fall apart.
$$\ce{ H2O(g) -&gt; 2H(g) + O(g)} \ \Delta H^\circ =+920\ \mathrm{kJ\cdot mol^{-1}}\ \Delta S^\circ =0.202\ \mathrm{kJ\cdot mol^{-1}\cdot K^{-1}}$$
By fixing $\Delta G=0$ at equilibrium, we can solve for a temperature at which this reaction becomes spontaneous:
$$T=\dfrac{\Delta H}{\Delta S}=\dfrac{+920\ \mathrm{kJ\cdot mol^{-1}}}{0.202\ \mathrm{kJ\cdot mol^{-1}}\cdot K^{-1}}=4596\mathrm{K}$$
We need to heat our water vapor up an additional 4218 K, which takes $18\ \mathrm{g}\times 1.996\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\times 4218\ \mathrm{K}=151.5\times 10^3 \mathrm{J}$.
So, we now have pumped nearly 200,000 J into our water sample, atomized it, and heated it to approximately 5000 K. We are now close to the temperature of the outer layers of the sun! Surely we have enough energy at this temperature to produce x-rays. Nope. At 5000 K, we produce minimal x-rays. Most of the radiation is in the visible, UV, and IR (think about what we get from the sun). Below is a plot of black-body radiation as a function of temperature (image by Wikipedia user Darth Kule and released into the public domain):

Okay, so we are far beyond the reality of what can happen in a conventional oven (or almost any reasonable heat source used for food). At this temperature, we can use the Planck Law to calculate the power output ($I$) of x-rays at the temperature. We can also do this at some normal temperatures and for gamma rays. This model is a little goofy, since food is not a black body, but we will at least calculate the max x-ray and gamma ray output.
Rather than grinding through all the maths, I'll just put in a table of some temperatures and watts. 1 watt is not a lot of power. Most lightbulbs produce light in the kilowatts.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{T\ (K)} &amp; \mathrm{P_{x-ray}\ (W)} &amp; \mathrm{P_{gamma}\ (W)} &amp; \mathrm{notes} \\ \hline
378 &amp; \approx 0 &amp; \approx 0 &amp; \text{boiling point of water} \\ \hline
550 &amp; \approx 0 &amp; \approx 0 &amp; \text{approximate common highest temperature on residential ovens}\\ \hline
700-800 &amp; \approx 0 &amp; \approx 0 &amp; \text{temperature range for wood-fired ovens, tandoors, etc.}\\ \hline
5770 &amp; 4.26\times 10^{-129} &amp; \approx 0 &amp; \text{temperature of the photosphere of the sun}\\ \hline
1.57\times 10^7 &amp; 10.4 &amp; 7.87\times 10^{-54} &amp; \text{estimated temperature of the center of the sun} \\ \hline 
\end{array}$$
So, if you could heat your food to the temperature of the sun, it would produce minuscule x-ray radiation. It would also no long resemble food.