This answer uses the terminology and notation of geometric algebra and geometric calculus.
A vectorial view of the Poisson bracket
The Poisson bracket of arbitrary scalar functions $G$ and $H$ of the canonical coordinates $(q_i,p_i)$ is normally written as
$$\{G,H\}=\sum_{i=1}^n \left(\frac{\partial G}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial G}{\partial q_i}\frac{\partial H}{\partial p_i}\right)\ .$$
Instead treating $G$ and $H$ as functions of the vectors $r=\sum q_i e_i$ and $p=\sum p_i e_i$, we can alternatively write the Poisson bracket as
$$\begin{aligned}
\{G,H\}&=(\partial_r G)\cdot(\partial_p H)-(\partial_p G)\cdot(\partial_r H)\\
&=[(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] G\ ,
\end{aligned}\tag{1}$$
where $\partial_r$ and $\partial_p$ are the vector derivatives with respect to $r$ and $p$.
Because $H$ is a scalar-valued function, the expression inside the $[\ ]$ on the RHS of (1) is just a scalar linear operator, which makes it straightforward to extend the definition of the Poisson bracket such that the left argument can be a multivector of any grade. In particular, with a vector $A(r,p)=\sum G_i(r,p) e_i$:
$$\begin{aligned}
\{A,H\}
&=[(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] A\\
&=\sum_{i=1}^n [(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] G_i e_i\\
&=\sum_{i=1}^n \{G_i,H\} e_i
\end{aligned}\tag{2}$$
Having thus recast the Poisson bracket in terms of vectors, we will no longer have any need to refer to components.
There are more powerful ways of expressing the Poisson bracket from a geometric algebra perspective than what we've done above1, but the above approach is adequate for the problem at hand.
Poisson bracket of the Laplace-Runge-Lenz vector and the Kepler problem Hamiltonian
The question involves the Hamiltonian
$$H=\frac{p^2}{2 m}- \frac{k}{|r|}$$
and the Laplace-Runge-Lenz vector, which is given in the question as
$$A=p\times (r\times p)-mk\hat{r}\ ,$$
where $\hat{r}=r/|r|$.
However, we're addressing the problem in terms of geometric algebra, which doesn't commonly use the cross product because it only works in three dimensions. So we twice use the identity
$$a\times b=-I a\wedge b\ ,$$
where $I$ is the unit pseudoscalar, to arrive at the more geometric algebra-friendly expression
$$A=-p\cdot (r\wedge p)-\frac{mk}{|r|}r\ .$$
Using the identity2
$$\partial_r \frac{1}{|r|}=-\frac{1}{|r|r^2}r\ ,\tag{3}$$
the vector derivatives of $H$ are seen to be the vectors
$$\partial_p H=\frac{1}{m}p\tag{4}$$
and
$$\partial_r H=\frac{k}{|r|r^2}r\ .\tag{5}$$
Note that in (2), it's actually the directional derivatives of $A$ that are needed, not the vector derivatives. Using (3) and multiple applications of the Leibniz rule, for an arbitrary vector $v$ the directional derivatives of $A$ are
$$(v \cdot\partial_r) A=-p\cdot(v\wedge p)-\frac{mk}{|r|}v+\frac{mk v\cdot r}{|r|r^2}r\tag{6}$$
and
$$(v \cdot\partial_p) A=-v\cdot(r\wedge p)-p\cdot(r\wedge v)\ .\tag{7}$$
Plugging (4)-(7) into (2) gives
$$\begin{aligned}
\{A,H\}&=\frac{1}{m}\left(-p\cdot(p\wedge p)-\frac{mk}{|r|}p+\frac{mkp\cdot r}{|r|r^2}r\right)\\
&\ \ \ \ -\frac{k}{|r|r^2}\left(-r\cdot(r\wedge p)-p\cdot(r\wedge r)\right)\\
&=\frac{k}{|r|r^2}\left(-r^2 p+(p\cdot r) r+r\cdot (r\wedge p)\right)\ .
\end{aligned}\tag{8}$$
But from the identity3
$$a\cdot(b\wedge c)=(a\cdot b) c-b(a\cdot c)$$
for vectors $a$, $b$ and $c$ we have that
$$r\cdot(r\wedge p)=r^2 p-r(r\cdot p)\ ,$$
which we can plug into (8) to give
$$\begin{aligned}
\{A,H\}&=\frac{k}{|r|r^2}\left(-r^2 p+(p\cdot r) r+r^2 p-r(r\cdot p)\right)\\
&=0
\end{aligned}$$
References
1: Geometric Algebra for Physicists, section 12.3
2: Clifford Algebra to Geometric Calculus, page 57, eq. 2.32
3: Ibid., page 12, eq. 1.42
