The classical electromagnetic waves may have many forms. The simplest examples are "monochromatic" waves with a well-defined frequency $f$ – the number of periods per second. The wavelength – the distance between two maxima of the wave – is $\lambda=c/f$.
When you change something about the electromagnetic field at one point, i.e. when you try to modify an electromagnetic wave, you can't instantly modify anything at a distant place. The information and influence can never propagate faster than $c$, the speed of light in the vacuum. This is a completely general fact about Nature that follows from the special theory of relativity and it applies to quantum theory, too. If you start with a wave of a huge wavelength $L$ and modify quickly something about the wave in a small region and in a timescale much shorter than the period of the electromagnetic wave, in a hope that this allows you to "instantly" affect distant points of space, you will fail. Instead of making a long electromagnetic wave collaborate on your project, you will create some electromagnetic waves of a much higher frequency and shorter wavelength. Too long electromagnetic waves just don't allow you to make things "really quickly" – in times shorter than $T$, their period – and too locally – in regions much shorter than the wavelength. Whenever the resolution of time or space is much better, it just proves that some electromagnetic waves of much higher frequency and shorter wavelength are present.
To describe the waves in the microwave oven, it is enough to consider classical physics i.e. avoid the notion of photons. The microwaves – electromagnetic waves whose wavelength is just a little bit shorter than the size of the oven – have a negligible chance to get through the holes because these holes are much smaller than the wavelength. The mechanism is sometimes called electromagnetic shielding. How does it work?
Well, if you want to discuss waves of wavelength $\lambda$ only, their dependence on space must always have the form $A\cdot \cos(2\pi x / \lambda)$: waves with the right distance between the minima. However, the metallic cage surrounding the interior of the oven imposes the potential $\phi=0$ at a very dense network of points. When you try to write down the potential as $A\cdot \cos(2\pi x / \lambda)$, while making sure that it's zero at all points where there is a conductor, you will find out that there is no solution except for $A=0$. The wave of the given wavelength just can't get through at all. Alternatively, you could calculate the reflection from the metallic points of the mesh (not counting the holes). They would interfere with each other and guarantee that the probability of reflection is nearly 100 percent.
In other words, the microwave-frequency photon is "really large", at least as large as the wavelength, and it just "doesn't fit in". It could fit in if it "pretended" it was a shorter-wavelength photon but that would be a different one. The microwave oven prescribes a frequency $f$ and the corresponding wavelength is always $c/f$ and cannot change. You may also look at the situation with a "poor resolution" so that you "neglect" distances shorter than the wavelength. From this viewpoint, the metallic caging of the oven looks solid despite the small holes. That's ultimately why it looks solid to the electromagnetic waves.
All these things may also be phrased in terms of photons – which is needed in quantum theory – even though, as I said, it is not necessary for microwaves because they contain a huge number of coherent photons so that this large group of photons behaves classically.
Individual photons are described by wave functions that mathematically look like an electromagnetic wave. It may be monochromatic but it may be a mixture of different frequencies, too. Individual photons' wave functions propagate through the oven or anything else pretty much just like the classical electromagnetic waves. That's why photons of microwave frequencies can't escape from the microwave, either. But the interpretation is different: the wave function isn't directly measurable like the electric fields. Instead, it encodes (after squaring) the probability density that the photon is here or there.
If you try to catch a photon whose wave function is spread over a large region, you have a certain calculable chance that you will catch it "somewhere here", in a particular nearby region $V$. But when you do so, the probability is zero that the photon is simultaneously at some other, distance place $P$. So whatever you do to the photon here – if you are lucky at all and the photon "shows up" here – will not affect what is happening very far from you. Effectively, if you see the photon here, the "wave function collapses" and you're sure that the photon isn't "over there" so it can't do anything there. The photon was never over there; it only had a chance to be there but your measurement showed that the chance didn't materialize. These points are often misunderstood because people are trying to imagine that the wave function of the photon is a real wave that must leave some traces even if the photon is ultimately seen elsewhere. But it can't and it doesn't leave any traces whatsoever: the nonzero wave function only quantified a "potential" that something may be seen somewhere. When it's seen elsewhere, it's a proof that the potential was never "real".
At any rate, you will see that photons can't be used to send superluminal signals, either.
The answer is this: You will need to cover the item you don't want to heat with a conducting material, creating a so called Faraday's cage, which is connected to an earth potential equal to the potential of the metal casing of the microwave oven.
The reason for this is threefold: The covering material needs to be a conductor, because conductors have the property that free charges can move free without significant resistance inside it. Because of this, the electric field inside the conductor is zero, for elsewise charges would move due to the electric field. But they move themselves in such a position that they cancel out the fiel again. Now, because the electric field inside the conductor material is zero, is also has to be zero in cavities of the conductor. For if it would not be zero inside the conductor, one could move electrons through a loop partially through the cavity, and partially through the conductor. The path integral through this closed loop would then be non zero. In onther words, one could generate energy without costs, obviously in contradiction to the law of conservation of energy. Therefore, wrapping an item in conducting material creates a so called Faraday's cage, whereby the electric field through the wrapped item is zero.
Now this is what we want, because the microwave oven heats items with the use of electric field waves. So, shielding the item off from electric fields, also shields it from heating.
However, charges inside the conductor start moving because of the electric field waves. This can create a potential difference with the shielding material of the casing of the microwave oven. It is because of this difference that arcing could occur. That is why you will also need te connect the conductive wrapping to the earth potential of the metal casing.
Best Answer
As John and others have said, the wavelength of the microwaves is very large compared to the size of the holes in the screen which allows the screen to act as a solid. Visible light has much smaller wavelengths and can pass through the holes unobstructed. It isn't possible to see (resolve) objects and features smaller than the wavelength of light (electromagnetic radiation) used so this is why the mesh works. See http://hyperphysics.phy-astr.gsu.edu/hbase/waves/mwoven.html for more details.