Apparently not. Let $\gamma=1/3$ and let $\varepsilon=\frac{1}{100}$.

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

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

Then for any subset $U\subseteq[0,1]$ and any measurable $B\subseteq\mathbb{R}^{k-1}$, we will have $\mu(A\Delta(U\times B))\geq\frac{1}{100}$.

Indeed, let $X_1=\left\{x\in[0,1];\quad\lfloor100x\rfloor\equiv0\text{mod}2\right\}$ and $X_2=[0,1]\setminus X_1$. Let $A_i=A\cap (X_i\times[0,1]^{k-1})$ and $U_i=U\cap X_i$ for $i=1,2$. Also let $C_1,C_2\subseteq[0,1]^{k-1}$ be the sets such that $A_1=(I_1\times C_1)\cup(I_2\times C_2)$, so that $\mu_{k-1}(C_1)=\mu_{k-1}(C_2)=\frac{1}{2}$ and $C_1,C_2$ are disjoint.

$$\mu(A\Delta(U\times B))=\mu(A_1\Delta(U_1\times B))+\mu(A_2\Delta(U_2\times B))=\mu((X_1\times C_1)\Delta(U_1\times B))+\mu((X_2\times C_2)\Delta(U_2\times B))$$
$$\geq\mu(X_1\times(C_1\setminus B))+\mu(X_2\times(C_2\setminus B))=\frac{1}{2}\mu([0,1]^{k-1}\setminus B).$$


So if $\mu(A\Delta(U\times B))<\frac{1}{100}$, we should have $\mu(B)\geq0.95$, thus $\mu(B\Delta C_i)>0.4$ for $i=1,2$, which implies
$$\mu(A\Delta(U\times B))=\mu((X_1\times C_1)\Delta(U_1\times B))+\mu((X_2\times C_2)\Delta(U_2\times B))
\geq
\mu(U_1\times(C_1\Delta B))+\mu(U_2\times(C_2\Delta B))
\geq0.4(\mu(U_1)+\mu(U_2))\geq0.4\mu(U)\geq0.4\mu(U\times B)\geq0.4(\mu(A)-\frac{1}{100})\geq0.4^2>\frac{1}{100},
$$
a contradiction.

Similar examples should work for $\gamma=1-\frac{1}{n}$, by letting $\varepsilon=\frac{1}{100n}$ and 
$$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100(n+1)x_i\rfloor\not\equiv0\text{ mod }n+1\right\}.$$