First, we may assume $(b_1,b_2,b_3)=1$.  That gives that for each $y\in (0,1)$ at most two of $\{b_1y,b_2y,b_3y \}$  can be integers.
Now, take $$J^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{3}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{3}}\right]$$ for all $1\leq k\leq b_i-1$. Assume $N$ to be large enough such that no two of these intervals (for all $i=1,2,3$ and all $k$) intersect each other, unless when they are both centered around the same point $y_0$, i.e both $b_iy_0$ and $b_jy_0$ are integers. If so, at most two of these intervals can intersect each other.

Also, define: $$I^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{2}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{2}}\right]\subset J^{(i)}_k.$$

On $I^{(i)}_k$ we have that $|u_i(y)|\sim N$, so if $y\in I^{(i)}_{k_1}\cap I^{(j)}_{k_2}$ then $|u_i(y)u_j(y)|\sim N^2$.But, for $l\neq i,j$ we have that $|u_l(y)|\ll N^\frac {1}{3}$ as $y\notin J^{(l)}_k$ for all $k$. Over a certain $I^{(i)}_k$ we can therefore bound $|u_1(y)u_2(y)u_3(y)| $ by $$|u_1(y)u_2(y)u_3(y)|\ll N^2N^\frac {1}{3}.$$ As $|I^{(i)}_k|=\frac{2}{b_iN^\frac{1}{2}}$, we have that:
$$\int _{I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^2N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{2}}=\frac{2}{b_i}N^2N^\frac {-1}{6}$$


There are at most $|b_i|$ such $I^{(i)}_k$'s, so the integral over all of them is bounded by $2N^2N^\frac {-1}{6}$.

Now, say we're in $J^{(i)}_k\setminus I^{(i)}_k$. Then at least one of the $u_i$'s is bounded by $N^\frac{1}{3}$, and the other two are at bounded by $N^\frac{1}{2}$. The length of each $J^{(i)}_k$ is $\frac{2}{b_iN^\frac{1}{3}}$, so:
$$\int _{J^{(i)}_k\setminus I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^\frac {1}{2}N^\frac {1}{2}N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{3}}=\frac{2}{b_i}N.$$
Again, the integral over the total of intervals $J^{(i)}_k\setminus I^{(i)}_k$ is bounded by $2N$.

Outside of all the intervals $J^{(i)}_k$, for all $i=1,2,3$ and $1\leq k\leq b_i-1$, we know that $|u_i(y)|\ll N^\frac {1}{3}$ and therefore $|u_1(y)u_2(y)u_3(y)|\ll N$. Everything is inside $(0,1)$, so the total of the integral "outside of the $J^{(i)}_k$'s" is bounded by $N$.

To summarize: $$\int _\frac {Q}{N}^\frac{1}{2}|u_1(y)u_2(y)u_3(y)|dy\ll 2N^2N^\frac {-1}{6}+2N+N=O(N^{1\frac {5}{6}}).$$

But, this is true only when we are far enough from 0. At 0, all three functions $u_i$ obtain their maximum simultaneously, so the main contribution to the integral will be near 0. take $b$ such that  $\frac{1}{bN^{\frac{1}{3}}}\leq\frac{1}{2b_{i}}$  for all $i=1,2,3$. Write:
$$\mathcal S= \left[\frac {Q}{N},\frac{1}{bN^{\frac{1}{3}}} \right].$$
 We estimate the integral on $\mathcal S$. Note that if $y\in \mathcal S$ then $b_{i}y\leq\frac{1}{2}$  and therefore $\left\Vert b_{i}y\right\Vert =\left|b_{i}y\right|$  for all $i=1,2,3$. Thus, $u_{i}\left(y\right)\ll\frac{1}{\left\Vert b_{i}y\right\Vert }=\frac{1}{\left|b_{i}y\right|}$  for $y\in\mathcal S$. Now:
$$\int_{\mathcal S}|u_1(y)u_2(y)u_3(y)|dy \ll \frac{1}{\left|b_{1}b_{2}b_{3}\right|} \int_{\mathcal S}\frac{1}{y^{3}}dy=O\left(\frac{N^{2}}{Q^{2}}\right).$$


We get that The integral of $u_1(y)u_2(y)u_3(y)$ on $\left[\frac{Q}{N},\frac{1}{2}\right]$ is $O\left(\frac{N^{2}}{Q^{2}}\right)$, and this bound is tight.