Why do larger black holes emit less Hawking Radiation than smaller black holes?

The Newtonian gravitational acceleration for an object of mass $M$ is given by the well known expression:

$$ a = \frac{GM}{r^2} \tag{1} $$

And the radius of the event horizon of a black hole is given by:

$$ r_s = \frac{2GM}{c^2} \tag{2} $$

Suppose we calculate the Newtonian gravitational acceleration at the event horizon. Let's not worry whether this is physically realistic for now, we'll just go ahead and do it anyway. If we rearrange equation (2) to get:

$$ GM = \frac{c^2r_s}{2} $$

then we can substitute for $GM$ in equation (1) to get:

$$ a(r_s) = \frac{c^2}{2r_s} \tag{3} $$

And what we find is that the gravitational acceleration at the event horizon is proportional to $1/r_s$ so smaller black holes have a higher surface acceleration than larger black holes. The temperature of the Hawking radiation is related to this surface acceleration. which makes intuitive sense. If the gravitational field is stronger you'd expect the Hawking radiation to be stronger. And that means:

Smaller black holes are hotter than larger black holes

To make this rigorous requires considerable work so I won't go into the details. We define a property called the surface gravity, $\kappa$, that is effectively the gravitational acceleration at the event horizon and we find for a static black hole this is given by:

$$ \kappa = \frac{1}{2r_s} \tag{4} $$

This is the proper general relativistic version of the Newtonian surface gravity I derived in equation (3). The Hawking temperature is then simply:

$$ T_H = \frac{\hbar c}{2\pi k_B} \kappa $$

So just as with our approximate calculation we find that the temperature is proportional to $1/r_s$ i.e. smaller black holes are hotter than larger black holes.

Here is an answer more based on thermodynamics. It all boils down to this: The entropy of a black hole is proportional to its surface area, but its mass = energy to the radius.

You can think of it in the following way: all states inside the black hole are the same to observers outside. Information about the interior is encoded on the surface because you can solve boundary value problems. So the larger the surface, the larger the entropy. This is far from a proof, but you can make this rigorous.

Now, what are heat and temperature? Heat is the spontaneous transfer of energy, without work being done. The second law of thermodynamics says that entropy can never decrease. Since energy is conserved, for two systems in equilibrium, $$\frac{\Delta S_1}{\Delta U} = \frac{\Delta S_2}{\Delta U}$$ that is, the the change in the entropy $S_1$ of system 1 due to changing its energy by $\Delta U$ is equal to the change in entropy $S_2$ of system 2 due to changing its energy by the same amount. Hence we define $$\beta = \frac{\partial S}{\partial U}$$ and you can realize that $\beta = 1/T$ where $T$ is the temperature. (Heat flows from low $\beta$ to high $\beta$, but from high $T$ to low $T$.)

From general relativity, we have that the mass $M$ and radius $r$ of a black hole are proportional, $r = 2GM/c^2$. If we take the energy to be $U = E = Mc^2$, then $U \propto r$, so $S \propto U^{2}$ which gives $\beta \propto U $ or $T \propto 1/r \propto 1/M$. Thus, a smaller black hole is hotter, and consequently radiates more.

If we invert the relation between entropy and energy, we get $U \propto \sqrt S$. Plot $\sqrt S$. It is very steep for small $S$, and quite flat for large $S$. That means for small black holes, the Hawking radiation only needs to have a little entropy, even if the intensity is large, for the process to be allowed by thermodynamics.

You can learn more from these talks by Hawking himself. Hawking uses some technical concepts, but I hope they are fairly accessible. (They are, at least, much more so than the original papers.)