This seems kind of trivially true (even without the assumption of $A$ Borel), unless I'm mistaken. First, let's see that $H(B_r(A)) \subset B_r(H(A))$. Any $y \in H(B_r(A))$ can be written as $\sum \lambda_i y_i$, with $y_1, \dots, y_m \in B_r(A)$, and $\lambda_1, \dots, \lambda_m$ non-negative reals adding to $1$. Let $x_i \in A$ be such that $||x_i - y_i|| \leq r$. Then it's clear from the triangle inequality that $||y - \sum \lambda_i x_i || \leq r$. So $y \in B_r(H(A))$, since $\sum \lambda_i x_i \in H(A)$. Conversely, let $y \in B_r(H(A))$. Then $||y-z|| \leq r$ for some $z = \sum \lambda_i x_i$ with $x_i \in A$ and $\lambda_i$ non-negative reals adding to $1$. Let $t = y - z$, and $y_i = x_i + t$. Then $||y_i - x_i || = || y - z|| \leq r$ and $\sum \lambda_i y_i = \sum \lambda_i x_i + (\sum \lambda_i) t = z + t = y$. That proves $y \in H(B_r(A))$.

This seems kind of trivially true (even without the assumption of $A$ Borel), unless I'm mistaken. First, let's see that $H(B_r(A)) \subset B_r(H(A))$. Any $y \in H(B_r(A))$ can be written as $\sum \lambda_i y_i$, with $y_1, \dots, y_m \in B_r(A)$, and $\lambda_1, \dots, \lambda_m$ non-negative reals adding to $1$. Let $x_i \in A$ be such that $||x_i - y_i|| \leq r$. Then it's clear from the triangle inequality that $||y - \sum \lambda_i x_i || \leq r$. So $y \in B_r(H(A))$, since $\sum \lambda_i x_i \in H(A)$. Conversely, let $y \in B_r(H(A))$. Then $||y-z|| \leq r$ for some $z = \sum \lambda_i x_i$ with $x_i \in A$ and $\lambda_i$ non-negative reals adding to $1$. Let $t = y - z$, and $y_i = x_i + t$. Then $||y_i - x_i || = || y - z|| \leq r$ and $\sum \lambda_i y_i = \sum \lambda_i x_i + (\sum \lambda_i) t = z + t = y$. That proves $y \in H(B_r(A))$.

Update: For the updated question (when you take intersections), there is indeed a counterexample. The first part of the proof still goes through, so the containment $H(B_r(A) \cap X) \cap X \subset B_r(H(A) \cap X) \cap X$ is still valid (since if the $y_i \in X$ their convex combination $y$ is too, because $X$ is convex). For a counterexample to the opposite inclusion consider the following picture in $\mathbb{R}^2$ (maybe someone can actually draw it :-) )

Let $p = (0,0)$, $q = (1,0)$ and $z = \frac{1}{2}(1, \tan(\theta))$ for some small angle $\theta$. Then $p,q,z$ form an isosceles triangle with angles $\theta, \theta, \pi - 2 \theta$. Let $X$ be this triangle (including the interior), and $A = \{p,q\}$. Then if $r < \tan(\theta)/2$ is small, the convex hull of $A$ is the line segment joining $p$ and $q$ which is contained in $X$, and so $B_r(H(A) \cap X) \cap X$ contains for instance the point $w = (1/2, r)$, since its distance from $(1/2,0)$ is $r$. But $B_r(A) \cap X$ consists of two disjoint small sectors of circles of radius $r$ and angle $\theta$ around $p$ and $q$. So the every point of it has $y$-coordinate at most $r \sin(\theta) < r$. So the same holds for its convex hull too. Hence $w$ is not in $H(B_r(A) \cap X) \cap X$. (On second thought, it seems that any $\theta < \pi/2$ will work.)