If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0<s<1$ the situation is a way more complicated, but the answer is "yes" if $H^{1+s}(B)<\infty$ and it follows from the following result due to Federer [F, Theorem 2.10.25], commonly known as the Eilebnerg inequality.

A metric space is boundedly compact if bounded and closed sets are compact.

Theorem. (Eilenberg inequality) Let $\Phi:X\to Y$ be a Lipschitz mapping between boundedly compact metric spaces. Let $0\leq m\leq n$ be real numbers (not necessarily integers). Assume that $E\subset X$ is $H^n$-measurable with $H^n(E)<\infty$. Then

• $\Phi^{-1}(y)\cap E$ is $H^{n-m}$-measurable for $H^m$-almost all $y\in Y$.

• $y\mapsto H^{n-m}(\Phi^{-1}(y)\cap E)$ is $H^m$-measurable.

Moreover $$ \int_Y H^{n-m}(\Phi^{-1}(y)\cap E)\, dH^m(y)\leq (\operatorname{Lip}(\Phi))^m \frac{\omega_m\omega_{n-m}}{\omega_n}\, H^n(E). $$

How take $\Phi:\mathbb{R}^2\to\mathbb{R}$, $f(x,y)=x$, $m=1$, $n=1+s$, $E=B$. If $H^{1+s}(B)<\infty$, then the function $$ x\mapsto H^{n-m}(\Phi^{-1}(x)\cap E)=H^{s}(B_x) $$ is $H^m$ measurable. Since $H^m=H^1$ is the Lebesgue measure, it is Lebesgue measurable.

The above argument is true under the assumption that $B$ is $H^{1+s}$-measurable with $H^{1+s}(B)<\infty$. If you know that $B$ is Borel, it is a much stronger condition than just measurability and I believe it is true without the assumption that $H^{1+s}(B)<\infty$, but I have no clue how to prove it and I do not even know if it is true. In fact I believe that the function $x\mapsto H^s(B_x)$ is measurable with respect to the $\sigma$-algebra generated by analytic sets. A comment that I learned from Pertti Mattila [EH, Remark 1.2] suggests that.

[EH] B. Esmayli, P. Hajłasz, The coarea inequality. Ann. Acad. Sci. Fenn. Math. (To appear).

[F] H. Federer, Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153 Springer-Verlag New York Inc., New York 1969.

If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0<s<1$ the situation is a way more complicated, but the answer is "yes" if $H^{1+s}(B)<\infty$ and it follows from the following result due to Federer [F, Theorem 2.10.25], commonly known as the Eilenberg inequality. See also [F, $\S$2.10.26] where the measurability of the integrand is addressed explicitly.

A metric space is boundedly compact if bounded and closed sets are compact.

Theorem. (Eilenberg inequality) Let $\Phi:X\to Y$ be a Lipschitz mapping between boundedly compact metric spaces. Let $0\leq m\leq n$ be real numbers (not necessarily integers). Assume that $E\subset X$ is $H^n$-measurable with $H^n(E)<\infty$. Then

• $\Phi^{-1}(y)\cap E$ is $H^{n-m}$-measurable for $H^m$-almost all $y\in Y$.

• $y\mapsto H^{n-m}(\Phi^{-1}(y)\cap E)$ is $H^m$-measurable.

Moreover $$ \int_Y H^{n-m}(\Phi^{-1}(y)\cap E)\, dH^m(y)\leq (\operatorname{Lip}(\Phi))^m \frac{\omega_m\omega_{n-m}}{\omega_n}\, H^n(E). $$

How take $\Phi:\mathbb{R}^2\to\mathbb{R}$, $f(x,y)=x$, $m=1$, $n=1+s$, $E=B$. If $H^{1+s}(B)<\infty$, then the function $$ x\mapsto H^{n-m}(\Phi^{-1}(x)\cap E)=H^{s}(B_x) $$ is $H^m$ measurable. Since $H^m=H^1$ is the Lebesgue measure, it is Lebesgue measurable.

The above argument is true under the assumption that $B$ is $H^{1+s}$-measurable with $H^{1+s}(B)<\infty$. If you know that $B$ is Borel, it is a much stronger condition than just measurability and I believe it is true without the assumption that $H^{1+s}(B)<\infty$, but I have no clue how to prove it and I do not even know if it is true. In fact I believe that the function $x\mapsto H^s(B_x)$ is measurable with respect to the $\sigma$-algebra generated by analytic sets. A comment that I learned from Pertti Mattila [EH, Remark 1.2] suggests that.

[EH] B. Esmayli, P. Hajłasz, The coarea inequality. Ann. Acad. Sci. Fenn. Math. (To appear).

[F] H. Federer, Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153 Springer-Verlag New York Inc., New York 1969.