There are many ways to carry heat.
The first is conduction, which is about the "vibration" of atoms on one material passing to another by simple physical contact. (Example: you touch something hot and get hurt).
The second is convection: hot molecules simply move from one place to another (Example, water starts to boil in the bottom of a pan, but moves on the top because is lighter).
The third is radiation and is precisely what you say: a warm body emits electromagnetic radiation. At "normal" temperatures (an oven, a human body), it's Infrared radiation, but it can be of higher frequency at higher temperatures, according to Planck's blackbody radiation law.
Notice, though, that the power emitted by radiation only is proportional to the fourth power of temperature. So the effect is very relevant in sun, but negligible for a human body. It should be around 500 W/m$^2$, which OK, is not small, but probably the most heat is transferred by conduction when the human is in air.
This is how garments work: they create a small layer of warm air around your skin, avoiding contact with constantly renewed cold air.
The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the object, not the environment. That equation makes some assumptions. I couldn't find a good explanation for why the coefficients are what they are in the net power formula you posted, so I thought I'd take a step back and derive it.
The power emitted per unit area from the surroundings is
$$P_s=\epsilon_s \sigma T_s^4$$
The object will absorb a fraction of that based on its area and absorptivity:
$$P_a=\alpha \epsilon_s \sigma T_s^4$$
The object will emit:
$$P_e=\epsilon \sigma T^4$$
The net power delivered to the object is
$$P_{net} = P_a - P_e = \epsilon\sigma T^4 - \alpha \epsilon_s \sigma T_s^4$$
If the absorptivity and emissivity are equal, and $\epsilon_s = 1$ (blackbody), we get:
$$P_{net} = P_a - P_e = \epsilon \sigma (T^4-T_s^4)$$
So you'd have to assume that the surroundings perfectly emitting, and that the absorptivity and emissivity are equal. The latter is true under thermodynamic equilibrium or local thermodynamic equilibrium. See the Wikipedia page for Planck's law and in particular the section on Kirchhoff's Law.
Best Answer
I'll answer your second question first, because then your first one is easier. In short, yes, the equilibration of temperature between two bodies is absolutely universal - it doesn't depend on how the heat is transferred, and in particular it does apply to radiative transfer. And, indeed, once two bodies have reached thermal equilibrium through radiative transfer, the radiation in between them has a temperature that is the same as the temperature of the two bodies.
Let us now imagine two bodies coming into radiative equilibrium. We'll say that one of them is a hollow sphere and the other one is inside it, because then we don't have to consider the surrounding environment (which I'll get to shortly). The inner body will be giving off heat at a rate proportional to its temperature to the fourth power, and this doesn't depend on the temperature of its surroundings at all. But it's also absorbing heat, at a rate that does depend on the temperature of its surroundings. (It's proportional to the fourth power of the outer body's temperature.) So when they come into equilibrium, the inner body is giving off and receiving heat at exactly the same rate. Radiation is still occurring, but heat flow is not, because the radiation coming in cancels the radiation going out, so there's no net flow of energy. This answers your first question.
Often we don't consider this because we're thinking about something that's a lot hotter than its surroundings. Because $T^4$ increases very rapidly with $T$, the radiative energy transfer from the environment is often small enough to be ignored in this case. In particular, for a body in space that's not exposed to sunlight, the relevant incoming radiation is the cosmic microwave background, which has a temperature of only 3 kelvin and can be ignored for most purposes.