Concentration inequality for function of independent Bernoulli r.v.'s (related to random graph)
In general, it seems that you cannot get the concentration you want without  more constraints on the probabilities. 

Consider the following graph on $n$ vertices. Let vertices $C = \{2,\ldots,n\}$ form a clique such that for all $i,j \in C$, $p_{ij} = 1$. Let vertex $1$ be connected to this clique via only one edge, with $p_{1,2} = 1/\log(n)$, and $p_{1,j} = 0$ for all $j \in C$ with $j \neq 2$. Then with probability $1/\log(n)$, all $n$ vertices are reachable from $1$ with 2 hops, and with probability $1 - 1/\log(n)$ none are. So the expected number of reachable vertices is $n/\log(n)$, but with probability $1/\log(n) \gg n^{-1}$, it is $n$, larger than its expectation by a $\log(n)$ factor. 

In fact, since you are interested in paths of length $\log n$, all that we require is that the realized edges in $C$ form an expander, not a clique, with high probability. This can be accomplished with $p_{ij} = o(1)$ as well, so this condition does not save you. 

The essence of the problem is this: since you can have a group of vertices $C$ that are highly connected with high probability, how many vertices vertex $1$ is connected to can depend only on whether it is connected to $C$ or not, which can depend on only a single edge. You can't expect something depending on only a single Bernouli random variable to be well concentrated.