On the number of integers with an even number of distinct prime factors Let $A(n)$ be the number of integers from 2 to $n$ with an even number of distinct prime factors, and let $B(n)$ be the number of integers  from 2 to $n$ with an odd number of distinct prime factors. (By "distinct" I mean, for example, that $2^{100}$ has one distinct prime factor.)

A little python program shows that as $n$ ranges from 2 to 1000000, there are only 9437 values of $n$ for which $A(n)&gt;B(n)$, whereas there are  990450 values of $n$ for which $A(n)&lt;B(n)$. There are 112 values of $n$ for which $A(n)=B(n)$, the largest of which is 12099.

Is there some precise theorem or conjecture or even plausible heuristic to the effect that $A(n)&lt;B(n)$ "most of the time", or perhaps "eventually"? Or is this just an illusion that dissolves as $n$ grows larger?
The Generalized Prime Number Theorem gives a decent estimate of the number of numbers with k factors less than n. You can use this to see that the number of numbers with 2,4,6,... prime factors should not in the large differ from the number of numbers with 1,3,5,... factors. 

The Generalized PNT states: 

$$\pi_k(x) \sim \frac{x}{\log x}\frac{(\log\log x)^{k-1}}{(k-1)!} $$

in which $\pi_k(x) $ is the number of numbers with k factors less than or equal to x.

See also problem 168307 (Generalized PNT in limit as numbers get large) which shows that in a sense even numbers have more prime factors than odd ones, and cites a result of Ramanujan that the "normal order" of prime factors of a number n, with or without repetitions, is about $\log\log n.$

For the last see Ramanujan, Collected Works, p. 274. For the Generalized PNT see, for example, G.J.O. Jameson, The Prime Number Theorem, p.145.

Edit in response to comment: This is similar to Chebyshev's bias and it could well be that generalized primes have similar properties. I don't think the proposition in the OP has been proven either way and even the proof of Chebyshev's bias is conditional.
The analysis using Landau's result was interesting but the error in Landau's result is $\sim O(\frac{1}{\log(\log(n)})$ and hence greater than the error derived $\sim (\frac{1}{\ln(n)^2})$, so unless we prove a tighter bound for k-almost primes than Landau did the rest of the argument won't apply.