Have to disagree with you on the bolded statement - they pass
RED Light - which is a VERY important qualifier
It's quite fundamental - the viewed light looks RED because that is the wavelength it is passing, whereas the (incandescent) source is made up of a broad spectra of wavelengths.
OK - if it looks pink, then admittedly it's not a perfect 'notch' filter* - but it's important to recognize that
PREDOMINANTLY it is the RED component of the spectrum that is being passed.
All well & good (sic!) for an incandescent - that is a broad spectrum light source of multiple wavelengths - the red filter allows the red wavelength to pass and blocks the others.
So yes, already it is quite inefficient for the emitted red light vs the total emitted light from the bulb
Now move to an LED - a White LED is NOT composed of the same spectral components as an incandescent; the red component is going to be SIGNIFICANTLY less.
The fact that is passes at all is because the red filter is not a perfect 'notch' filter.
The bottom line is that a RED Filter passes RED Wavelength light - the highest percentage of total total light that is passed through a RED filter is going to be from a RED LED - even if the total emitted raw lumens is less - because ALL of the light is going to be a much narrower wavelength bandwidth, much closer to matching the red filter.
* The wavelength of a RED LED is VERY narrow - certainly MUCH better than the bandwidth of the filter
So the filter is going to pass virtually all of the light (OK there may be some attenuation but that is going to be small relative to the non-RED blocking)
Some reading material -
Standard and White LED Basics and Operation - Application Note - Maxim
i.e. there is NO Red Wavelength component in a White LED - the only reason it emits through the LENS at all, is because the LENS itself is not very focused on a narrow RED wavelength.