Theorem 3.2 in http://arxiv.org/pdf/1405.2593.pdf shows that for any $x$ there are $\gg x\exp(-\sqrt{\log x})$ integers $x_0 \in [x; 2x]$ such that $\pi(x_0 + \log x) - \pi(x_0) \gg \log\log x$.
Is there an upper bound for number of such $x_0$? I think it must be $<x(\log x)^{-c}$ for any $C$.
UPD: It is interesting to find such upper bound for prime numbers $x_0$ such that $\pi(x_0 + \log x) - \pi(x_0) \gg \log\log x$
If $x_0<x$ satisfies that $[x_0, x_0+\log x]$ contains $\log\log x$ primes, then for a parameter $r$ we have that this interval contains $\binom{\log\log x}{r}$ different $r$-tuples $p, p+d_1, p+d_2, \ldots, p+d_{r-1}$ of primes, such that $0<d_1<\dots<d_{r-1}\leq\log x$. The number of possible choices for $d_1, \ldots, d_{r-1}$ is $\binom{\log x}{r-1}$. Apply Selberg's sieve to each of them, and take the sum over all tuples. This will lead to some lengthy computation involving singular series, but on average the singular series will be of magnitude $\mathcal{O}(1)$. From this you obtain that the number of short $r$-tuples is bounded above by $$\ll C(r)\binom{\log x}{r-1} \frac{x}{\log^r x} \ll \frac{C(r)}{r!}\frac{x}{\log x},$$ where $C(r)$ is the coefficient of Selberg's sieve. Each tuple belongs to $\leq\log x$ values of $x_0$, hence comparing with the $\binom{\log\log x}{r}$ tuples produced by a single $x_0$ and restricting $x_0$ you obtain that there are $\ll C(r)\frac{x}{(\log\log x)^r}$ possibilities for $x_0$. Optimizing for $r$ should give an upper bound of magnitude $\frac{x}{(\log x)^c}$ for some $c>0$, which falls somewhat short of your expectation.
Addendum: Since $r$ is large, it is better to use the large sieve in place of Selberg's sieve. The details will become more complicated, but the results should be better. For a model you can look at the proof of Lemma 2 in Elsholtz, On cluster primes, Acta Arith. 109, 281-284.
You might assume that since the chance of a random integer p being prime is roughly p / ln p, the number of primes in an interval of fixed width n gets smaller at smaller. And you might make a conjecture that no n+1 consecutive positive integers contain more primes than the n+1 integers from 2 to 2+n.
We call a sequence of increasing integers x_k with 0 <= x_k <= n a "prime pattern of length n" if for every p the set of values x_k modulo p has fewer than p elements. There is the conjecture that for every prime pattern there are infinitely many primes p such that p + x_k is prime for every x_k. For example (0, 2) is a prime pattern because there is only one value modulo 2, and the conjecture is the twin prime conjecture.
It seems that there is no prime pattern of length n containing more numbers than the integers from 2 to n+2 contain primes. For example for n = 5, there are 4 primes 2, 3, 5 and 7, but there is no prime pattern of four numbers with length < 8, and prime patters of five numbers are even longer.
However, an exhaustive search shows that for n around 2200 there are prime patterns containing more numbers than the number of primes from 2 to n+2, and therefore it is conjectured that there is an interval of length about 2,200 containing more primes than the integers from 2 to n+2. And it is likely that this is true for arbitrary large n, so for every n there are intervals p to p+n containing more primes than the interval 2 to n+2.
The worst case is if the primes are spread out, so that there is at least (log x)/2 between each pair of primes (and log log x > 2), so the best case (if you admit reasoning by analogy) is if the primes are spaced in an interval at a density of loglog x primes per log x of interval. This gives an upper bound on your quantity of x/log log x. Hopefully someone who knows the literature can give a reference to a better upper bound.
