Apparently not. Let $\gamma=1/3$ and let $\varepsilon=\frac{1}{10}$.

For any $k\in\mathbb{N}$ we can consider the set 

$$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100x_i\rfloor\equiv0\text{mod}2\right\}.$$
(So, $A$ is the `white squares' in a $k$-dimensional chess board of side $100$)

Then for any interval $I\subseteq\mathbb{R}$ and any measurable $A_0\subseteq\mathbb{R}^{k-1}$, we will have $\mu(A\Delta(I\times A_0))>\varepsilon$.

Indeed, we obviously have $\mu(A\Delta(I\times A_0))>\varepsilon$ if the interval $I$ has length $<\frac{1}{10}$ because then almost all the set $A$ lies outside $I\times A_0$, and if $I$ has length $>\frac{1}{10}$ then $\frac{\mu(A\cap(I\times A_0))}{\mu(I\times A_0)}\leq0.6$ (this can be seen by comparing the intersections of $A$ and $I\times A_0$ with each interval $[0,1]\times p$, for each $p\in[0,1]^{k-1}$).

Similar examples can be constructed for $\gamma=1-\frac{1}{n}$, by letting $\varepsilon=\frac{1}{10n}$ and 
$$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100(n+1)x_i\rfloor\not\equiv0\text{ mod }n+1\right\}.$$