Here is a winning strategy for Olya. Denote by $L_n$ the total length of the white segments after the $n$-th round ($L_0$ is considered to be equal to 1).
Statement. Olya can play in such a way that for $k = 0, 1, \dots, 1012$ after the $(2k)$-th round the total length of the white segments is greater than $\frac{1}{2}$ ($L_{2k} > \frac{1}{2}$), and for some $\delta > 0$ at least one of the open intervals $]0, \delta[$ or $]1 - \delta, 1[$ is completely white.
Let us prove this statement by induction.
The basis for $k=0$ is obvious.
Assume that after the $(2k)$-th round the situation is as described in the statement. Consider Tolya's move in the $(2k+1)$-th round. If he chooses $l=0$ or $l=1$, then Olya can choose the same $l$ on her turn in the round $2k+2$, and the segment $[0, 1]$ will not change its color, with the exception, perhaps, of two points in the case $l=0$.

Now suppose that Tolya chose $0 < l < 1$. Since at least one of the outermost open intervals is white, Olya can choose $J$ in such a way that after the $(2k+1)$-th round both outermost open intervals $([0, \delta]$ or $]1-\delta, 1[$ for some $\delta > 0$) were painted white. If $L_{2k+1} > \frac{1}{2}$, then Olya simply chooses $l = 0$. Let's consider the case when $L_{2k+1} \le \frac{1}{2}$. Since the outermost open intervals of the interval $[0, 1]$ are white, there exists $\epsilon > 0$ such that the intervals $]0, \epsilon[$ and $]1-\epsilon, 1[$ are white. Then if Olya chooses the number $l \in ]1-\epsilon, 1[$, for example $l = 1-\frac{\epsilon}{2}$, then whatever segment $J$ Tolya chooses, he will completely cover all black intervals and leave one or two white intervals with a total length of $\epsilon$. That is, $L_{2k+2} = (1-L_{2k+1}) + \epsilon > \frac{1}{2}$. It is also clear that Tolya will not be able to repaint both outermost intervals black at once.