This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.
Lemma 1: Suppose that $0 < r < 1$. Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}_{\ge 0} \rbrace$. Fix $k \ge 1$. Let $S_k$ be the set of sums of the form $s_1+\cdots+s_k$ such that $s_i \in S$ and $|s_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum_{i \in I} s_i = 0$. Then $0$ is not in the closure of $S_k$.
Proof: Use induction on $k$. The base case is trivial: $S_1=\lbrace -1,1\rbrace$. Now suppose $k \ge 2$. If a sequence $(x_i)$ in $S_k$ converges to $0$, then the smallest summand in the sum giving $x_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S_k$, which are all nonzero. Discarding the finitely many $x_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x_i$ yields a sequence $(y_i)$ in $S_{k-1}$ tending to $0$, contradicting the inductive hypothesis.
Now fix $b>1$ and $k$. Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}_{\ge 0} \rbrace$. Let $T_k$ be the set of sums of the form $t_1+\cdots+t_k$ with $t_i \in T$.
Lemma 2: Each $t=t_1+\cdots+t_k \in T_k$ equals $u_1+\cdots+u_\ell+\delta$ for some $\ell \le k$ and some $u_i \in T$ with $u_i = O(t)$ and $\delta = O(1)$.
Proof: Examine the powers of $b$ used in the $t_i$. If any nonempty subsum (with signs) of these powers equals $0$, the corresponding $t_i$ sum to $O(1)$. If $b^n$ is the largest power that remains after removing all such subsums, divide all the remaining $t_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t_i$, which are $O(b^n)$, are $O(t)$.
Corollary: The number of elements of $T_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.
Corollary: $T_k \ne \mathbb{Z}$.
We have
$$(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$
$$= (\operatorname{sgn}(i) \cdot2^j)^{-1} \int_0^1 \big(x^{l+2} (1-x)^{2(j+2)}\big)/(1+x^2)\; dx +\frac{k}{ |i|\cdot2^j }\int_0^1x^l(1-x)^{2(j+2)}\; dx $$
Now $\frac{1}{2^j }\int_0^1x^l(1-x)^{2(j+2)}\; dx $ is just some rational number so by selecting $k$ and $i$ we can take the second term to be any rational number (positive if $k$ is required to be positive).
So if we evaluate the left term and find it to be $\pm \pi$ plus a rational number, we can deduce that for any other number of the form $\pm \pi$ plus a rational number (with the same sign in the $\pm$) we can choose $k,i$ to make the integral equal that formula.
This is easy to do. If we set
$$\frac{x^{l+2} (1-x)^{2j+2}}{1+x^2} = \frac{a+bx}{x^2} + f(x)$$
for $a,b$ two rational numbers and $f(x)$ a polynomial in $x$ with rational coefficients, then
$$x^{l+2} (1-x)^{2j+4} = a+bx +(1+x)^2 f(x)$$
$$ i ^{l+2} (1-i)^{2j+4} = a + bi$$
$$a+bi = i^{l+2} (-2i)^{j+2} = 2^2 \cdot 2^j \cdot i^{l-j} $$
As long as $j$ or $l$ are congruent modulo $2$, this is $\pm 4 \cdot 2^j$, so the leftmost term is the integral of a polynomial with rational coefficients plus
$$ \pm 4 \int_0^1 \frac{1}{1+x^2} dx = \pm \pi$$ as desired.
Best Answer
Using the idea of the other answer in a different way, if $u$ is the irrationality measure of $\pi$, then except for finitely many $p/q$, we have
$$ \left| \pi - \frac{p}{q} \right| > \frac{1}{q^u} $$
and consequently
$$ \frac{ -\log |\pi - (p/q)| }{\log p + \log q } < \frac{u}{2}$$
and there will be infinitely many fractions $p/q$ that come arbitrarily close to this bound. (and, of course, those finitely many exceptions which may exist that are allowed to exceed it) (and, whatever tiny excesses might arise due to the rounding error in the analysis)
If the irrationality measure if $\pi$ is greater than $2.34$, then there will be infinitely many fractions with a better value of $R$ than the one you found. (although that is not reason to expect any of them are small enough for us to actually find)
If $u < 2.3$, then there can only be finitely many fractions with a better value of $R$. But I have no idea how you would go about checking if any exist at all.
Almost all irrational numbers have irrationality measure $2$; for $\pi$ it's known that $u \leq 7.6063$