Solution:

Ross can't stop Josie winning — Josie has a strategy in which she can ensure that there will be 8 white stones in a row. We will give an explicit example of such a strategy.

To simplify notation, we define a $k$-strip to be a $1 \times 8$ rectangle, in which the first $k$ cells are filled with white stones and the other $8 - k$ cells are empty.

- Step One. Josie creates 32 disjoint 1-strips using the following technique.

Start by finding 44 disjoint $1 \times 9$ rectangles on the board.

- Josie places a white stone in one end of each of these 44 rectangles on her first 22 turns.
- Ross "ruins" 22 of them. Each of the other 22 rectangles are 1-strips (by ignoring the empty end cell).
- On Josie's next 10 turns she then chooses 20 of the ruined $1 \times 9$ rectangles, and "unruins" them by placing a white stone in the other end. These are now 1-strips (by ignoring the cell containing the black stone).
- However Ross "ruins" a further 10 of them. So in total we have
$$22 + 20 - 10 = 32 \text{ disjoint 1-strips.}$$

- Step Two. Repeat the following operation for $k = 1, 2, 3, 4, 5$.

Starting with $2^{6 - k}$ disjoint $k$-strips. Josie can use her next $2^{5 - k}$ turns to add one white stone to each of these strips. On Ross' next $2^{5 - k}$ turns he can spoil at most $2^{5 - k}$ of them. So we are left with at least $2^{5 - k}$ disjoint $(k + 1)$-strips.

- Step Three. Now we have at least one 6-strip. Josie wins immediately by placing both her white stones into the empty cells in the 6-strip.