Do most bitstrings expand if they halt when executed by a Universal Turing machine?
Both can happen.

The counting argument applies to all TMs that always halt.

In the partial world, you can construct Turing machines for both extreme cases (and everything in between) like so. Assume $T$ is a Turing machine. Construct

T&lt;(x) = if |T(x)| &lt; |x| then return T(x) else loop


and

T&gt;(x) = if |T(x)| &gt; |x| then return T(x) else loop


Note that if T(x) loops, both T&lt;(x) and T&gt;(x) loop; that's not a problem.
It's easy so see that for T&lt;, all halting inputs contract, and for T&gt; all expand.

Rice's theorem immediately shows that these expansion/contraction properties (and related ones) are undecidable, by the way.