Does a low-pass filter average the higher-order frequencies into a "DC offset"?
To answer the specific question in your title: no, a low pass filter does not “average higher order frequencies” it attenuates them.

To answer your actual question: yes, the basic procedure of demodulation and filtering will do what you want. 

The math is rather straightforward. Simply multiply by \$ \sin(\Omega t) \$ and integrate over one period and you will see the term you want. Given your equation, using \$ \sin(\Omega t) + \cos(\Omega t)\$  instead as the multiplier should produce a cleaner result. 

The issue is in how to implement the integration in a way that does not compromise your feedback loop stability. The simplest way is to use a low pass filter with a cutoff frequency of at most \$ \Omega  / 2\$ to ensure the removal of feedthrough and double-frequency components. 

Do note that a real-life continuous integrator is always just a low-pass filter. An actual integrator has infinite gain at DC, real components with real finite gains and leakage will limit this gain to perhaps \$ 10^5 \$ if not less, which directly implies a low-pass response.

Of course, much lower cutoff frequencies are commonly used, so as to have low-order practical filters. How much attenuation is necessary will depend on your application, but in general you would use filter that has a bandwidth not much higher than that of your signal of interest (in this case your feedback loop response speed). 

You can also design a filter with specific zeros at the problematic frequency components (\$ \Omega \&amp; 2\Omega \$). But in most cases, a very simple low pass is much more than enough.