Yes, Braker can guarantee that any two boxes will contain different number of balls. Suppose that Writer at the first move writes a pair $(A_1, A_2)$. At each move Braker chooses pairs containing box $A_1$ (if possible). By doing that he can guarantee that all pairs $(A_1, A_i)$, $i = 2,3,\ldots,2018$ are on the table. Braker numerates all $2015$ pairs not containing $A_1$ randomly by numbers $1,2,\ldots,2015$ and accordingly puts balls into these boxes. Let $t(A_i)$ be the total number of balls in the box $A_i$ after this procedure. Without loss of generality, assume that
$$t(A_2) \le t(A_3) \le \dots \le t(A_{2018})$$
After that Braker for each $i = 2,3,\ldots,2018$ numerates $(A_1, A_i)$ by $2014 + i$ and accordingly distributes balls. Hereby we get
$$t(A_2) < t(A_3) < \dots < t(A_{2018}).$$
For each $i = 2,3,\ldots,2018$ the box $A_i$ received balls at most $2016$ times and only in one case the number of balls was more than $2015$. Therefore,
$$t(A_1) = 2016 + 2017 + \dots + 4032 > 2016 \cdot 2016 + 4032 > t(A_i)$$
and as a result any two boxes contain different number of balls.