Not true.  For example, if you have a 100 watt CW transmitter
with the key held down, the carrier power is 100 watts.
Now, what happens when you send a string of dots (50% duty)?
Your total power out is indeed 50 watts, but only 25 watts
is in the carrier; the other half is in the keying sidebands
(if ideal envelope shaping).  In general, the carrier power
is multiplied by the square of the duty cycle.

But doesn't the lost power depend on the Morse speed? Are you saying that for a dot lasting an hour, 50% of the total key down power is taken up by the few milliseconds at key up/down?

No.  If you send one-hour dots every two hours, and had a
"key click" filter with a time constant of an hour, then
a 100-watt Tx at 136 kHz would put out 25 W on 136000 Hz,
12.5 W on ~ 136000.000139 Hz, and 12.5 W on ~ 135999.999861 Hz.
If you use a filter with a time constant of a few milliseconds,
the carrier power is still only 25 W.  However, there are now
also sidebands at ~ 136000.000417 Hz, 136000.000694 Hz, etc.
In either of the above cases, nearly all the sideband energy
would fall within your receiver's bandwidth, so you could
indeed claim 50 W average useful output from the Tx.

However, a LORAN blanker would be gating the signal on and
off much faster than the desired signal's message modulation.
The sidebands created would be offset from the carrier by

10 Hz, and would not even appear on the Argo screen.

--Stewart