Can a photon have a wavelength less than the planck length? Well? Maths if necessary, please. 
Would the photon effectively have no wavelength?
 A: The short answer is we simply do not know: this hypothesis is utterly beyond anything that we can either test experimentally or reason about with a widely accepted theory.
Симон Тыран's Answer is a good, concise exposition showing what classical, relativistic reasoning has to say about this. But I don't believe this answers the question  because you have to make the assumption that classical relativity works down to an arbitrarily small scale and that's something we neither know nor (I get the impression as a lay reader) even believe. 
Moreover, even without full quantum gravity, one can formulate classical theories with both an invariant velocity (as in the $c$ of special relativity) and and invariant length scale. Examples of such theories are "Doubly Special Relativity" and also de Sitter Invariant Special Relativity wherein the symmetry group $SO(1,\,4)$ is a supergroup of the Lorentz group and is the same as the symmetry group of de Sitter space, a highly symmetric vacuum solution of the Einstein field equations. In such a universe one would have a natural, invariant length scale that can be used to invalidate the classical special relativistic reasoning that there always exists an inertial observer whom a wavelength is arbitrarily small for.
A: We don't know if massless particles may have a wavelength beyond planck length, but there are some significative indices (including the ones provided in the other answers) against limitation of wavelength. In particular, according to Wikipedia,

There is currently no proven physical significance of the Planck
  length; it is, however, a topic of theoretical research.

This research is referring in particular to the discreteness of spacetime and not to wavelengths.
The possibility to increase the relative velocity of an observer is an essential argument, and it is important to note that the spacetime interval of massless particles is zero. A zero interval means that there is no place for  a n y  kind of wavelength, even not for a wavelength smaller than Planck length. However, contrary to space intervals, the spacetime interval does not correspond to any observer but to a (hypothetical, non existent) observer moving at speed of light. You can now assume an observer moving very close to speed of light, and as far as we know today there is no limit, that means the observer will never reach speed of light, but he may come arbitrarily close to it, and this observer would observe a wavelength which may be arbitrarily small, that means arbitrarily close to the zero length of the zero spacetime interval of the massless particle.
Now, the speed of the observer is a space interval per time interval, and by this, your question transforms into the question if admitted wavelengths are discrete, in particular if space and/ or time are discrete. But as I mentioned above, mainstream today considers space and time to be continuous.
A: If there was a minimum wavelength, you just could increase your own velocity in direction of the photon to still make it smaller. 
Since the relative velocity to the photon must always be c the only thing that can increase then is the frequency, so you always can get a still smaller wavelength just by increasing the doppler effect. 
Because of that there is no minimum wavelength. If you could reach c (which you can't) the wavelength would be zero, but since you can get arbitrarily close to c the wavelength can also get arbitrarily close to zero.
A: I am going to weigh in on this. In one sense we can say that we really do not know. Since you mentioned experimental evidence, we have no experimental evidence of anything at the Planck energy $E~=~\sqrt{c^5/G\hbar}$. The wave length of a particle at this energy would be equal to the Planck length. The Planck length is computed by equating the wavelength of a particle at rest with the circumference of the event horizon. The $4$-momentum $P^\mu~=~(mc,~0,~0,~0)$ with the deBroglie type equation $p\lambda~=~h$, we equation $\lambda$ equated to $2\pi$ times the Schwarzschild radius $r~=~2GM/c^2$ the rest is algebra and you get $\ell_p~=~\sqrt{G\hbar/c^3}$
So what does it mean for a particle to have a wavelength shorter than $\ell_p$? It means the particle is in a region smaller than the even horizon of the smallest quantum unit of black hole. There is nothing immediate that says this can't happen. What we do hypothesize is that this unit of black hole represents the smallest region one can locate a qubit. 't Hooft and Susskind formulated holography initially by looking at event horizons according to units of Planck areas that can hold a qubit of information. 
So suppose you have photons or any quantum field with an arbitrary spectrum of energy or frequencies. The energy in a Fourier sum that exceeds the Planck energy is not able to hold a qubit of information. In other words, these probably do not play any physically meaningful role and can be removed. This follows in some sense with the idea that the Planck scale is sort of the ultimate renormalization cut-off in QFT or quantum gravity. Of course as yet the details of this are not yet certain.
