Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$ groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$.

Here is Lubotzky's paper: https://core.ac.uk/download/pdf/82373534.pdf

Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$-groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$.

Here is Lubotzky's paper:

Enumerating Boundedly Generated Finite Groups, Journal of Algebra 238 (2001) pp 194–199. doi:10.1006jabr.2000.8650, core.ac.uk version.

Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. However, this isn't quite good enough. The number of $2$ groups of order $n = 2^m$ is $n^{B \log n}$ where

$$B = \frac{2}{27 \log 2}.$$

You would win if $B > A$, but Lubotzky doesn't compute $A$, and a quick look at the argument suggests that the bound is not going to be in your favor. Looking at papers which cite Lubotzky's paper I do not quickly see any better bounds in the literature, so you may be out of luck.

Here is Lubotzky's paper: https://core.ac.uk/download/pdf/82373534.pdf

Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$ groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$.

Here is Lubotzky's paper: https://core.ac.uk/download/pdf/82373534.pdf