Let $d\in(0,2R)$. We will construct by induction
a family of affine subsets $\{P_n\subset H:n\in\mathbb{N}\}$
two sequences of vectors $\{e_n\in P_n:n\in\mathbb{N}\}$, $\{c_n\in P_n:n\in\mathbb{N}\}$
a sequence of reals $\{r_n:n\in\mathbb{N}\}\subset \mathbb{R}_+$,
such that
$ S (c_n,r_n)\cap P_n= S (0,R)\cap P_n\tag{1}$
$P_n\subset P_{n-1}\tag{2}$
$e_n\in S (c_n,r_n)\cap P_n\tag{3}$
$\Vert x-e_{n-1} \Vert=d\tag{4}$
for all $n\in\mathbb{N}$ and $x\in P_n\cap S (c_n,r_n)$.
Properties $(1)-(4)$ will guarantee that $\Vert e_k-e_l\Vert=d$ for $k\neq l$. Indeed, without loss of generality we assume $l>k$. Conditions $(1)$ and $(2)$ gives $ S (c_n,r_n)\cap P_n\subset S (c_{n-1},r_{n-1})\cap P_{n-1}$. Hence using $(3)$ we get $e_l\in S (c_l,r_l)\cap P_l\subset S (c_n,r_n)\cap P_n$. Then by $(4)$ we conclude that $\Vert e_k-e_l\Vert=d$. Therefore we may take $A:=\{e_n:n\in\mathbb{N}\}$.
Let $c_1=0$, $r_1=R$ and $P_1=H$, then choose arbitrary $e_1\in S (c_1,r_1)\cap P_1$. Assume we have already constructed sequences $\{c_1,\ldots,c_n\}\subset H$, $\{e_1,\ldots,e_n\}\in H$ and $\{r_1,\ldots,r_n\}\subset\mathbb{R}_+$ satisfying $(1)$,$(2)$ and $(3)$. Consider affine function $f_n(x)=\langle x-c_n, e_n-c_n\rangle$ and define $ P_{n+1}=\{x\in P_n: f_n(x)=r_n^2-d^2/2\} $ $ c_{n+1}=c_n+\left(1-\frac{d^2}{2r_n^2}\right)(e_n-c_n) $ $ r_{n+1}=d\sqrt{1-\frac{d^2}{4r_n^2}} $ Later you'll see why, but now just see this picture 
From definition of $P_{n+1}$ we see $P_{n+1}\subset P_n$, so condition $(2)$ satisfied.
For arbitrary $x\in S (c_{n+1}, r_{n+1})\cap P_{n+1}$, we have $x\in P_{n+1}\subset P_n\subset\ldots\subset P_1=S (0,R)$, so $x\in S (0,R)\cap P_{n+1}$. This gives inclusion $ S (c_{n+1}, r_{n+1})\cap P_{n+1}\subset S (0,R)\cap P_{n+1}$. Now let $x\in S (0,R)\cap P_{n+1}$, then $\Vert x\Vert=R$ and $f_n(x)=r_n^2-d^2/2$. Moreover since $x\in P_{n+1}\subset P_n$nd $x\in S (0,R)$ we have $x\in S (0,R)\cap P_n$. Recall that $ S (0,R)\cap P_n= S (c_n,r_n)\cap P_n$, so $x\in S(c_n,r_n)$ and $\Vert x-c_n\Vert^2=r_n^2$. Since $e_n\in S(c_n,r_n)\cap P_n$, then $\Vert e_n-c_n\Vert^2=r_n^2$. It is remains to recall definitions of $c_{n+1}$ and $r_{n+1}$ to get $ \Vert x-c_{n+1}\Vert^2= \Vert x-c_n\Vert^2+\Vert c_{n+1}-c_n\Vert^2-2\langle x-c_n, c_{n+1}-c_n\rangle= $ $ \Vert x-c_n\Vert^2+\left(1-\frac{d^2}{2r_n^2}\right)^2\Vert e_n-c_n\Vert^2-2\left(1-\frac{d^2}{2r_n^2}\right)\langle x-c_n, e_n-c_n\rangle= $ $ \Vert x-c_n\Vert^2+\left(1-\frac{d^2}{2r_n^2}\right)^2\Vert e_n-c_n\Vert^2-2\left(1-\frac{d^2}{2r_n^2}\right)f_n(x)= $ $ r_n^2+\left(1-\frac{d^2}{2r_n^2}\right)^2 r_n^2-2 \left(1-\frac{d^2}{2r_n^2}\right)\left(r_n^2-\frac{d^2}{2}\right)=d^2\left(1-\frac{d^2}{4r_n^2}\right)=r_{n+1}^2 $ i.e. $x\in S(c_{n+1},r_{n+1})$. Also we know $x\in P_{n+1}$, so $x\in S(c_{n+1},r_{n+1})\cap P_{n+1}$. Since $x\in S (0,R)\cap P_{n+1}$ is arbitrary we get inclusion $S (0,R)\cap P_{n+1}\subset S(c_{n+1},r_{n+1})\cap P_{n+1}$. Both inclusions gives $S (0,R)\cap P_{n+1}= S(c_{n+1},r_{n+1})\cap P_{n+1}$, hence condition $(1)$ is satisfied.
As the consequence $ S (c_{n+1},r_{n+1})\cap P_{n+1}= S(0,R)\cap P_{n+1}\subset S(0,R)\cap P_n=S(c_n,r_n)\cap P_n $.
Now take arbitrary $x\in \ S (c_{n+1},r_{n+1})\cap P_{n+1}$. Since $x\in P_{n+1}$ then $f_n(x)=r_n^2-d^2/2$. Since $x\in S(c_{n+1},r_{n+1})\cap P_n\subset S(c_n,r_n)\cap P_n$, then $\Vert x-c_n\Vert=r_n^2$. Since $e_n\in S(c_n,r_n)\cap P_n$, then $\Vert e_n-c_n\Vert^2=r_n^2$. Thus $ \Vert x-e_n\Vert^2=\Vert x-c_n\Vert^2+\Vert e_n-c_n\Vert-2\langle x-c_n, e_n-c_n\rangle=r_n^2+r_n^2-2f_n(x)=d^2 $ Hence for all $x\in S (c_{n+1},r_{n+1})\cap P_{n+1}\subset P_n$ condition $(4)$ holds.
Now take arbitrary $x\in S (c_{n+1}, r_{n+1})\cap P_{n+1}$ and set $e_{n+1}\in x$. By the choice condition $(3)$ holds.
Since all conditions are satisfied we constructed by induction the desired sequences. But in fact this induction may breaks down if at some step $r_n$ becomes a complex number. Thus we need to study when the recurrence defined by $ r_{n+1}=d\sqrt{1-\frac{d^2}{4r_n^2}} $ and $r_1=R$ will exist. Denote $x_n=r_n/d$, then we get recurrence $ x_{n+1}=\sqrt{1-\frac{1}{4x^2}} $ with $x_1=R/d$. Since $d\in(0,2R)$, then $x_1\in(1/2,+\infty)$. Now take a look a this graph.
Here we make plots of functions $x$ and $\sqrt{1-\frac{1}{4x^2}}$. We see that they touches at the point $x=2^{-1/2}$
We see that this recurrence is infinite iff $x\geq 2^{-1/2}$, otherwise this is finite. In terms of $d$, this means that our recurrence well defined iff $d\leq R\sqrt{2}$.
Unfortunately we conclude that this method doesn't provide a way to get an infinite set of elements with pairwise distance between elements equal to $d>R\sqrt{2}$. It seems to mee that there is no such sequence. But anyway one can slightly modify formula for $r_{n+1}$ and $c_{n+1}$ to get the sequence of elements with pairwise distance not equalt to the same value but still greater than $R\sqrt{2}$.