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 $$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100(n+1)x_i\rfloor\not\equiv0\text{ mod }n+1\right\}.$$

Apparently not. Let $\gamma=1/3$ and choose some $\varepsilon<\frac{1}{4}$.

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

$$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\;\sum_{i=1}^k\lfloor2x_i\rfloor\equiv0\text{mod}2\right\}.$$ (So, $A$ is the `white squares' in a $k$-dimensional chess board of side $2$)

Then for any subset $U\subseteq[0,1]$ and any measurable $B\subseteq\mathbb{R}^{k-1}$, we will have $\mu(A\Delta(U\times B))\geq\frac{1}{4}$. This follows from the proof of the claim below and the fact that $A$ can be expressed (up to measure $0$) as $([0,1/2)\times C_1)\cup([1/2,1)\times C_2)$, where $C_1,C_2\subseteq[0,1]^{k-1}$ are disjoint and have measure $1/2$.

Claim: Let $A:=[0,1/2)^2\cup[1/2,1)^2\subseteq[0,1]^2$. For any subsets $X_1,X_2$ of $[0,1]$, $\mu((X_1\times X_2)\Delta A)\geq\frac{1}{4}$.

Proof:

Let $x:=\mu([0,1/2)\cap X_1),y:=\mu([1/2,1)\cap X_1)$, $z:=\mu([0,1/2)\cap X_2),w:=\mu([1/2,1)\cap X_2)$.

Clearly $\mu((X_1\times X_2)\Delta A)$ only depends on $x,y,z,w$, and in fact (as shown in the picture, where $(X_1\times X_2)\Delta A$ is shaded) we have $\mu((X_1\times X_2)\Delta A)=\frac{1}{2}+xw+yz-xz-yw$.

Thus, the problem is reduced to minimizing the function $xw+yz-xz-yw$ for $x,y,z,w\in[0,1/2]$, and one checks that the minimum is $-\frac{1}{4}$, concluding the proof.

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

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 $$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100(n+1)x_i\rfloor\not\equiv0\text{ mod }n+1\right\}.$$

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 $$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\quad\sum_{i=1}^k\lfloor100(n+1)x_i\rfloor\not\equiv0\text{ mod }n+1\right\}.$$