Ok, so as a rule:
$$Watts_{in} = Watts_{out}$$
This is always true unless you're storing energy somewhere.
A 400W heater will produce 400W of thermal energy.*
A 400W incandescent will produce 400W of thermal energy and light energy combined. Unfortunately if you measure the amount of light energy being produced, it is rather small compared to the total energy consumption of 400W.
A 400W LED fixture will produce 400W of thermal energy and light energy combined, but the amount of light energy is going to be a more significant figure in the process. Now you will be able to tell that it's not 400W being lost to the heat sink, but some amount less than that. (This ratio is the efficiency). The rest is being emitted as mostly-visible light.
However, if you put the LED in a well-insulated black box, you would not be able to tell the difference between a 400W lightbulb and a 400W LED by the temperature of the box.
In general, the actual efficiency of a light source (in Watts emitted as light) is not published.
Efficacy is not the same as efficiency. It takes into account the human eye response that determines how bright the light is, as produced from the source. This is not in units of Watts/Watt like efficiency, but in Lumens/Watt.
For example, comparing a white LED to a UV or deep blue LED, the white LED would have much higher efficacy compared with the deep blue LED, even if the amount of energy released in photons is the same, because the human eye does not respond to deep blue very well.
*As an aside, a 400W heat pump can present more than 400W of thermal power at its output, but it does that by moving heat rather than purely producing it.
Best Answer
Pedantically, assuming we are driving each of these 1 watt LEDs to their full power handling capability, they put out exactly the same radiant flux: 1 watt, or 1 joule per second. Then again, so does a 1W resistor.
The reason: with 1W of electrical power going in, there must be at equilibrium 1W of power going out. Some of it will be red light, and the rest of that electrical power will go into heating the component until blackbody radiation is enough to achieve thermal equilibrium. Within the normal operating temperatures of these components, that blackbody radiation will be in the far infrared spectrum, but the definition of radiant flux includes all electromagnetic radiation, even that which is not visible.
The LEDs will emit less photons than the resistor, because at least some of their radiation will be at a higher frequency. Higher frequency photons have more energy, so fewer are required.
But probably, you care just about the visible radiation from the LEDs. To do this, you have to define just what you mean by "visible". A standard definition is luminous flux:
For sure, either of your 1W LEDs will have a much higher luminous flux than a resistor. The 630nm LED has a bit of an edge over the 660 nm LED as the eye is more sensitive to 630nm (peak sensitivity being somewhere around 550 nm). However, this is but one of many variables needed to answer your question. The particulars come down to the specifics of the device. Read the datasheets.
To address some of your additional concerns:
Less energy per photon, but this is really irrelevant for everyday purposes.
$$\text{energy per photon} = h\nu = \frac{hc}{\lambda} $$
Where:
See Energy vs. power in transmitters from Amateur Radio Stack Exchange. Although that's about radio transmitters, it's still electromagnetic radiation just like visible light (but at a much lower frequency)
Assuming for this comparison an equal radiant flux. You've got it backwards: higher wavelength means a lower frequency, thus less energy per photon. To attain a given radiant flux (power), you can have:
being emitted each second. The number of photons emitted per second, multiplied by the energy per photon equals the radiant flux:
$$ \text{photons per second} \cdot \text{energy per photon} = \text{radiant flux} $$
If this is the case, and you want to weight all radiation in this visible range equally, you can do that. In this case, there's really very little difference besides color between a 660nm and 630nm LED (or an LED of any other color). Any differences in efficiency by this definition will be a consequence of the device's construction. For example, the resistance of the leads, the shadowing of the LED die by the bond wires, the opacity of the lens, flaws in the manufacture, etc.
No. Although a 630nm photon and a 660nm photon have different energies, the rate they are emitted will be different such that the radiant flux will be identical.
Think of it is this way: if you have a factory that consumes raw water at 100 liters per second and has to ship that water out at the same rate, you can:
Each photon is a package of energy. The LED consumes electrical energy at some rate (1W), and it must output some other kind of energy at the same rate (probably emitting photons, either as visible light or blackbody radiation). It doesn't matter what size the packages are.