Update: Pending independent verification, the answer to the title question is "no", according to a computation of $q(10) = 11609679812$ (which is even).
Let $q(n)$ be the number of ones in the binary expression for $10^{10^{n}}$ (which is the same as for $5^{10^{n}}$ of course), for positive integer $n$. The first nine $q(n)$ values — the most I can compute — are all odd: $11, 105, 1163, 11683, 115979, 1161413, 11606847, 116093517, 1160951533$. Is this just a quirk (with even values showing up eventually), or is $q(n)$ odd for all positive integer $n$? Equivalently, letting $t(0),t(1),t(2)...$ be the Thue-Morse sequence, is $t(10^{10^{n}}) = 1$ for all positive integer $n$?
It may not be surprising that about half of the binary digits of $5^{10^{n}}$ are ones, i.e., $q(n) \sim \displaystyle\frac{1}{2} \log_2(5^{10^{n}}) \approx \ 1.16096404744 \cdot 10^{n}$ (thus explaining the leading digits of sufficiently large terms), but why should all the terms be odd (if indeed they are)?
Note: My motivation for posting is the expectation that a proposition like "$q(10^{10})$ is even" provides a problem of the kind that Solomon Feferman ("Are there absolutely unsolvable problems?", p. 16), calls "absolutely unsolvable from the standpoint of practice". Obviously, this expectation is wrong if it can be proved that $q(n)$ is odd for all positive integer $n$. I strongly suspect, however, that there are infinitely many $q(n)$ values of each parity.