The problem is that the statement is a little imprecise, so it has some truth and some falsity.
A photon is, by definition, a quantum of energy exchanged between the elecromagnetic field (more precisely: a mode of the electromagnetic field) and something else. So by definition the energy exchanged is proportional to the number of photons, and therefore the light intensity is proportional to the number of photons exchanged through a surface, divided by the surface area and the time of the exchange. We speak of "quanta" of energy because the electromagnetic field cannot give or receive energy continuously, but only in multiples of a minimum amount – the quantum, the photon.
But this isn't the full story. The energy of a photon depends on the frequency $\nu$ of the electromagnetic field exchanging it: the energy is $h\nu$, where $h$ is Planck's constant. So if the same number of photons is exchanged in two different situations, through equal areas and in equal amounts of time, the intensity can still be different in the two situations.
How does the temperature enter in all this? Roughly speaking, in many situations we're actually uncertain about the number of photons exchanged by the electromagnetic field. We only have a probability that a particular number is exchanged. Typically this probability depends on the temperature $T$. For example (thermal state of light) we can have that the probability that $n$ photons of frequency $\nu$ are exchanged is
$$P(n)=\frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ ,$$
where $k$ is Boltzmann's constant.
This means that the precise value of the intensity is also uncertain, because the number of photons is. We can calculate the average or expected value of the intensity: by definition this average is proportional to
$$\sum_n n\ P(n) \equiv \sum_n n\ \frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ .$$
To summarize, the intensity is proportional to the number of photons, but the probability of the exact number of photons depends on the temperature. Therefore the average intensity depends, indirectly, on the temperature. It's also important to specify whether we're speaking about the temperature at one specific frequency (or of one specific electromagnetic mode), or through the whole spectrum, and so on.
It's good to emphasize that notion of temperature in electromagnetism is tricky. From a classical thermodynamic point of view the electromagnetic field does not have any temperature. Rather, we speak of the temperature of the matter in which it resides or with which it interacts. From a statistical-mechanical point of view we can speak of the (statistical) temperature $T$ of the electromagnetic field, in the sense that if we're uncertain about its state we then give each possible state a probability proportional to $\exp[E/(kT)]$ (or some other Gibbsian formula), where $E$ is the state's energy. This is almost a definition of the (statistical) temperature of the field. This definition goes over to the quantum description, and that's where the formula for the probability given above comes from. The two notions of temperature get connected only through the presence of matter: in equilibrium the statistical temperatures of the electromagnetic field and of the matter it interacts with must be the same; but the statistical temperature of matter is in turn connected with its thermodynamic temperature (in ways that are still under debate).
Good references about all this are Leonhardt's Measuring the Quantum State of Light (Cambridge U. Press 1997); Jackson's Classical Electrodynamics (Wiley 1999); Hutter, van de Ven, Ursescu's Electromagnetic Field Matter Interactions in Thermoelastic Solids and Viscous Fluids (Springer 2006); and also Biró's Is There a Temperature?: Conceptual Challenges at High Energy, Acceleration and Complexity (Springer 2011).
