First note that there are $8$ possible colors for each cell ($7$ rainbow colors plus white), hence there are $8^2 = 64$ three-celled rectangles with the two outer cells having the same color and $\frac{1}{2}\times 8^2\times 7 = 224$ with two outer cells having different colors. This means there are exactly $288$ possible distinct rectangles - so every rectangle must appear exactly once. For each ($1\times 3$) rectangle $R$, let $o(R)$ and $c(R)$ denote the number of white cells in $R$'s two outer cells and $R$'s center cell respectively. (so $o(R)\in\{0,1,2\}$ and $c(R)\in\{0,1\}$ always.) Since all colors appear equally, $\sum_{R} o(R)=\frac{1}{8}(288\times 2)=72$ and $\sum_{R} c(R)=\frac{1}{8}(288)=36$. Now note that both sums can also be counted from the white cells. The $48$ white cells adjacent to $Q$ contribute $48$ to $\sum_{R} o(R)$ and $0$ to $\sum_{R} c(R)$. (since they appear in the $48$ rectangles that extend outside $Q$ - all as 'outer' cells) Therefore the white cells in $Q$ must contribute $24$ to $\sum_{R} o(R)$ and $36$ to $\sum_{R} c(R)$. However, as $24<36$, this is impossible from the fact that each white cell $w$ in $Q$ contribute exactly $2$ to $\sum_{R} c(R)$ and at least $2$ to $\sum_{R} o(R)$ (that is, there are two rectangles with center $w$ and $\geq 2$ with $w$ as an outer cell) Thus the answer is: Petya will not succeed.

First wo prove that every rectangle must appear exactly once. This part is the same as talkon's solution. Assume the colors $c_1, \dots ,c_7, W$. Then we have a look at the rectangles: $c_1c_2c_1, \dots, c_1c_2c_7, c_1c_2W$(probably rotated). Obviously, any cells except the $W$ can't lie outside the $12*12$ square. So there must be a cell next to the $c_1$ on the extended line of $c_2c_1$. Like this: $A_1c_1c_2c_1A_2, A_3c_1c_2c_2, \dots, A_9c_1c_2W$. And there must be two cells among $A_1, \dots, A_9$ having the same color. Then there are two rectangles with the form $Xc_1c_2$ painted the same. Contradiction. So Petya won't succeed.