I haven't been able to find a reference that addresses this specific point, but I'm farily sure that "uniformly close to the center" has the second meaning from the question, i.e.
$$
\sup_{x \in A} \frac{|x - c_{B_x}|}{r_x} < 1.
$$
Here's an adaptation of a proof of the centered Besicovitch covering lemma from some notes that handles the uncentered case for balls.
Step 0: Parameters
We begin by defining a few parameters that will be important as we go along. First, let
$$
\gamma = \sup_{x \in A} \frac{|x - c_{B_x}|}{r_x},
$$
so that $\gamma < 1$. Select $\alpha$ satisfying $(1 + \gamma)/2 < \alpha < 1$, and further define $\delta = (\alpha - \gamma) / (\alpha + 1)$, so that $0 < \delta < 1$. Finally, let $R = \sup_{x \in A} |x|$.
Step 1: Selecting the cover
Now we iteratively select a sequence of balls from our collection. Define $R_1 = \sup_{x \in A} r_x$. We can assume that $R_1$ is finite, as otherwise we can find a single ball in the collection which covers $A$. (Since we're allowing uncentered balls, the uniformity condition comes into play here.) Pick the first ball $B_1 = B(c_1, r_1)$, associated with the point $x_1 \in A$, so that $r_1 > \alpha R_1$. Having selected $B_1, B_2, \ldots, B_m$ (and assuming they do not cover $A$), let
$$
R_{m+1}
= \sup_{x \in A \setminus \bigcup_{k=1}^m B_k} r_x
$$
and select $B_{m+1} = B(c_{m+1}, r_{m+1})$ associated to $x_{m+1} \in A \setminus \bigcup_{k=1}^m B_k$ so that $r_{m+1} > \alpha R_{m+1}$. Note that for $i < j$, we have
$r_i > \alpha R_i \geq \alpha R_j \geq \alpha r_j.$
The balls $\{ B'_k = B(c_k, \delta r_k) \}$ are then disjoint: assuming $i < j$, we have $x_j \notin B(c_i, r_i)$ and as a result
\begin{align*}
|c_i - c_j|
&\geq |c_i - x_j| - |x_j - c_j| \\
&\geq r_i - \gamma r_j \\
&> \delta r_i + (1 - \delta) \alpha r_j - \gamma r_j \\
&= \delta r_i + \delta r_j.
\end{align*}
We can also say that every $B'_k$ is contained in the ball $B(0, 2R_1 + R)$ since $|c_k| \leq |c_k - x_k| + |x_k| \leq R_1 + R$. It then follows that the radii $r_k \to 0$, and so $R_k \to 0$ as well. We can in turn conclude that the main balls $\{ B_k = B(c_k, r_k) \}$ cover $A$, as otherwise we obtain a contradiction to $R_k \to 0$.
Step 2: Bounding intersections, small radii
Going forward, we let $M$ denote a (potentially large) parameter whose value we defer selecting until later. (Its value will ultimately depend on the values of $\gamma$ and $\alpha$, but otherwise there won't be more specific dependence on the starting collection of balls.) We now show that for any $m$, there are at most $(2M + 1)^n \alpha^{-n} \delta^{-n}$ of the balls $B_k$, $k = 1, 2, \ldots, m-1$ which both intersect $B_m$ and satisfy $r_k \leq M r_m$.
To see this, fix an $m$ and suppose that $B_{k_1}, \ldots, B_{k_{N}}$ satisfy the given criteria. Then the associated dilated balls $B'_{k_1}, \ldots, B'_{k_{N}}$ are disjoint (as showed in Step 1) and are further contained in $B(c_m, (2M+1) r_m)$: for $x \in B(c_{k_i}, \delta r_{k_i})$ we have
\begin{align*}
|x - c_m|
&\leq |x - c_{k_i}| + |c_{k_i} - c_m| \\
&\leq \delta r_{k_i} + r_{k_i} + r_m \\
&\leq 2 M r_m + r_m.
\end{align*}
Measure considerations now give
\begin{align*}
(2M + 1)^n r_m^n
\geq \delta^n \sum_{i = 1}^{N} r_{k_i}^n
\geq \delta^n \sum_{i = 1}^{N} \alpha^{n} r_m^n
= N \alpha^{n} \delta^n r_m^n,
\end{align*}
since $r_{k_i} > \alpha r_m$ as showed in Step 1. This establishes the claimed bound.
Step 3: Bounding intersections, large radii
Next suppose that $i < j < m$ and that the balls $B_i$ and $B_j$ both intersect $B_m$, while also $r_i, r_j > M r_m$. We establish a lower bound for the angle made by the rays from $x_m$ to each of $c_i$ and $c_j$.
Since $x_m \notin B_i$ by construction, we first have the lower bound $| x_m - c_i | \geq r_i$, and likewise $| x_m - c_j | \geq r_j$. We also have the upper bound
$$
| x_m - c_i |
\leq | x_m - c_m | + | c_m - c_i |
\leq (1 + \gamma) r_m + r_i,
$$
and likewise replacing $i$ by $j$. Third, we have a lower bound
\begin{align*}
| c_i - c_j |
&\geq | c_i - x_j | - | x_j - c_j | \\
&\geq r_i - \gamma r_j \\
&\geq (\alpha - \gamma) r_j \\
&> 0,
\end{align*}
relying on the fact that $x_j \notin B_i$.
Letting $\theta$ denote the angle between the rays, the law of cosines now gives
\begin{align*}
\cos \theta
&= \frac{|x_m - c_i|^2 + |x_m - c_j|^2 - |c_i - c_j|^2}{2 |x_m - c_i| |x_m - c_j|} \\
&\leq \frac{[(1 + \gamma) r_m + r_i]^2 + [(1 + \gamma) r_m + r_j]^2 - [r_i - \gamma r_j]^2}
{2 r_i r_j} \\
&= \gamma
+ \frac{1 - \gamma^2}{2} \frac{r_j}{r_i}
+ (1 + \gamma)
\left(
\frac{r_m}{r_i} + \frac{r_m}{r_j}
\right)
+ (1 + \gamma)^2 \frac{r_m}{r_i} \frac{r_m}{r_j}.
\end{align*}
We further have $r_j / r_i < 1 / \alpha$ since $i < j$, and by hypothesis both $r_m / r_i$ and $r_m / r_j$ are less than $1/M$. Therefore
$$
\cos \theta
\leq \gamma
+ \frac{1 - \gamma^2}{2 \alpha}
+ \frac{2 (1 + \gamma)}{M}
+ \frac{(1 + \gamma)^2}{M^2}.
$$
Our initial choice of $\alpha$ ensures that
$$
\gamma
+ \frac{1 - \gamma^2}{2 \alpha}
< 1,
$$
and so by now selecting $M = M(\gamma, \alpha)$ sufficiently large we can also bound $\cos \theta$ away from $1$, thereby bounding $\theta$ away from $0$. That is, we can say that there is an $\epsilon = \epsilon(\gamma, \alpha)$ such that $\theta \geq \epsilon$.
To conclude this step, suppose that $m$ is fixed and that among the balls $B_k$, $k = 1, 2, \ldots, m-1$, there are $N$ which both intersect $B_m$ and satisfy $r_k > M r_m$. The angular separation condition just established then implies that there is a constant $C = C(n, \gamma, \alpha)$ such that $N \leq C$.
Step 4: Wrapping up
Putting together Steps 2 and 3, we can say that there is a constant $C' = C'(n, \gamma, \alpha)$ so that for any $m$, at most $C'$ of the balls $B_1, B_2, \ldots, B_{m-1}$ intersect $B_m$.
Finally, it follows that
$$
\sum_{k} \chi_{B_k}(x) \leq C' + 1
$$
for all $x$. For otherwise, we can select $x$ and $m$ so that $x \in B_m$ and
$$
\sum_{k = 1}^{m} \chi_{B_k}(x) > C' + 1.
$$
So then $x$ lies in more than $C'$ of the balls $B_1, \ldots, B_{m-1}$, meaning that more than $C'$ of these balls intersect $B_m$.
EDIT:
Here's a 1-dimensional counterexample to show that a Besicovitch-type result isn't ensured by the condition $\sup_x |x - c_{B_x}| < \infty$.
Let $A = (0, 1)$, and for $x \in A$ let $B_x = (0, (1+x)/2)$. Note that $|x - c_{B_x}| \leq 1/2$ for every $x$. It's straightforward to see that no finite collection of these balls (intervals) covers $A$. But the interval $(0, 1/2)$ is contained in each of the balls, so if $A' \subseteq A$ is such that $\{ B_x : x \in A' \}$ covers $A$ then $\sum_{x \in A'} \chi_{B_x}$ is infinite on $(0, 1/2)$.
