Solution:

Because $2011\equiv11\pmod{1000}$, it is enough to find $11^{2011}=(10+1)^{2011}$ modulo $1000$.

All terms with $10^3$ or higher are divisible by $1000$, so only the first three terms matter: $$1+2011\cdot10+\binom{2011}{2}10^2.$$

Modulo $1000$, this is $1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100\equiv1+110+500=611$.

The hundreds digit is therefore $6$.

Thus, D is the correct answer.