Yes, you *can *assign a temperature to the radiation from the Sun. It is approximately a blackbody radiation field with a temperature of 5800 Kelvin. But no, it doesn't "cool" on the way to the Earth.
The radiation field is still a blackbody radiation field with a temperature of 5800K when it reaches the Earth, but its intensity (or power per unit area) has decreased from $6.4 times 10^{7} Wm^{-2}$ at the surface (photosphere) of the Sun, to about $1.4 times 10^3 Wm^{-2}$, when it reaches the Earth.
It is this latter figure that is important in terms of how much the Earth is heated, not the temperature of the radiation field, because it determines how much energy per second is fed into the Earth's biosphere. An analogy would be an electric bar heater. The element will be emitting radiation at a temperature of 1000 C, but you don't burst into flames if you sit more than a metre away from it, because the power is distributed over a larger area.
In the absence of an atmosphere, the Earth would achieve an equilibrium temperature, such that it radiated away as much energy as it absorbed. This equilibrium temperature is much lower than the temperature of the Sun's surface because the power per unit area incident on the Earth is much less than the power per unit area emitted at the Sun's surface. The calculation, which I won't bore you with can be found here.
$$ T = T_{odot}(1-a)^{1/4}(R_{odot}/2D)^{1/2},$$
where $T_{odot}$ is the temperature of the Sun, $R_{odot}$ is the radius of the Sun and $D$ is the distance between the Earth and the Sun. $a$ is the "albedo" - the fraction of the Sun's light that is reflected back into space.
Putting numbers - $a=0.3$, $L_{odot}=3.8times10^{26} W$, $R_{odot}=7times10^{8} m$, $D=1.5times 10^{11} m$; we find $T= 256 K$ or $-16 C$. Of course the presence of an atmosphere warms us over this simple calculation by some tens of degrees.
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