I think the answer is no. Call the opposite of two points the point such that the three points together form an equilateral triangle. It is Vasya's turn after 33 points are colored. He colors the 34th point such that the two outermost points are the same color. Let the opposite of these two points be A. Then, no matter which side Petya plays on, Vasya plays on the other side such that it is the same color as the colored point 33 points away. Call the opposite of these two points B. A and B are adjacent, so Vasya will be able to color either A or B, and form an equilateral triangle.

Is this work? The answer is negative. Label point $P_1, P_2,...,P_{99}$. WLOG Petya color $P_1$ to red. On the second turn, Vasya paint $P_{99}$ to red. Now Vasya will paint in such a way that the figure is symmetric along perpendicular bisector of $P_1P_{99}$. Clearly this will ensure that $P_{17}$ and $P_{83}$ are the same color. Vasya will follow this plan until either the point $P_{49}$ or $P_{51}$ is painted. Then she will paint $P_{50}$ to the same color as $P_{17}$ and $P_{83}$, creating equilateral triangle so Vasya wins.

Nice problem! Solved with Mathnerd2007 ARO 2018 G9 P5 wrote: On the circle, 99 points are marked, dividing this circle into 99 equal arcs. Petya and Vasya play the game, taking turns. Petya goes first; on his first move, he paints in red or blue any marked point. Then each player can paint on his own turn, in red or blue, any uncolored marked point adjacent to the already painted one. Vasya wins, if after painting all points there is an equilateral triangle, all three vertices of which are colored in the same color. Could Petya prevent him? Answer is NO i.e. Vasya will win no matter what. First 2 observations, 1. The parity of Petya's moves are odd (i.e. Petya moves $1st,3rd,5th,.....$) and that of Vasya is even. 2. The indices of the vertices of an equilateral triangle are of the form $K,K+33,K+66$. We will show that we will get an equilateral triangle on the $98th$ move i.e. Vasya's $49th$ move. The Strategy is as follows, Consider the game after the $33rd$ move, say the coloured points(consecutive) are $(K,K+1,K+2,.....,K+32)$ and the next move is of Vasya's. On this move Vasya will colour point $(K+33)$ with the same colour as that of point $K$.Then Vasya will play symmetrically with Petya i.e. if Petya colours $K+98$(Indices taken modulo $99$) Vasya will colour $K+34$ with the same colour,i.e. if Petya colours $K+I(I>33)$ Vasya will just colour $K+132-I$ with the same colour. Now since the number of points between $K$ and $K+66$ and between $K+33$ and $K+66$ is $32$,note that on the $97th$ move Petya will have to colour $K+65/K+67$ if Vasya follows the aforementioned strategy . Thus on the $98th$ move Vasya will colour $K+66$ with the same colour as that of $K+33,K$ and get the equilateral triangle ($K,K+33,K+66$).This concludes OUR Proof $\blacksquare$