We will play an auxiliary game. If $n$ is odd, let $a=\frac{2^n-2}{3}$, $b=\frac{2^n+1}{3}$ and $c=\frac{2^n+4}{3}$. If $n$ is even, let $a=\frac{2^n-1}{3}$, $b=\frac{2^n+2}{3}$ and $c=\frac{2^n+5}{3}$. We start the game with a pile of $a$ red balls with value $\frac{1}{3a}$, $b$ green balls with value $\frac{1}{3b}$, and $c$ blue balls with value $\frac{1}{3c}$. Every minute, Petya splits the remaining pile into two piles of distinct value, and Vasya removes one of them. Petya wins if after $n$ minutes the pile is non-empty and all balls have the same color. By considering the value of the pile at each step, a winning strategy for Petya in the auxiliary game translates into a winning strategy in the original game. We will now give a winning strategy. Let $S$ be a pile of balls. We will say that the pile is almost balanced (AB) in different cases, depending on the value of $|S|$ modulo 3. If $|S|=3k$, the color distribution must be $(k-1, k, k+1)$. If $|S|=3k+1$, the color distribution must be $(k, k, k+1)$. If $|S|=3k+2$, the color distribution must be $(k, k+1, k+1)$. Observe that the initial distribution is AB on either $2^n+1$ or $2^n+2$ balls. An easy case analysis shows the following. An AB pile on $2^k$ balls can be split into two AB piles of $2^{k-1}$ balls, an AB pile on $2^k+1$ balls can be split into an AB pile on $2^{k-1}$ balls and an AB pile on $2^{k-1}+1$ balls. And if $k$ is even, an AB pile on $2^k+2$ balls can be split into two AB piles on $2^{k-1}+1$ balls. Moreover, the value of the two new piles in each case is distinct. By playing like this, we can ensure by induction that after $k$ minutes the remining pile of balls is AB on either $2^{n-k}$ or $2^{n-k}+1$ balls. After $n-1$ minutes, we will either have two balls of different colors, or three balls where two balls have the same color and the third one has a different color. In either case, we can win in one move.

Discussed with EthanWYX2009. Sketch of solution, any mistakes are mine alone. The main idea is Claim. $\frac14$ can be represented in $4$ ways as a sum of $2^{n-2}$ fractions with numerator $1$, each way using different fractions from the others with denominator less than or equal to $2^n + 50$ Proof. I'm annoyed, the idea is to use $4$ of the numbers among $\frac{1}{2^n + 1}$,$\frac{1}{2^n + 2}$, $\frac{1}{2^n + 4}$, $\frac{1}{2^n + 16}$, $\frac{1}{2^n + 32}$ in the $4$ representations, details need to be filled. After that label each number used in the same representation with $A$, $B$, $C$ or $D$, and notice we can put them in $4$ groups equidistributing each label in the $4$ groups, we may assume that not all are equal (as we may exchange some numbers of same label but different value), therefore, we can pair them and avoid get two equal numbers, and no matter what pair of groups gets delete we can repeat the same process, until the $n-2$-th step, where we will have numbers of labels $A$, $B$, $C$, $D$, we can still pair them as to avoid equality, and at the end the fraction remaining will of some label, that by the main claim have denominator at most $2^n + 50$. $\Box$ I fake solved initially by using a naive representation and thinking that it works.