Firstly, we calculate the units digit of the sum of squares from $1$ to $10$: $$ 1^2 + 2^2 +\cdots + 9^2 + 10^2 \equiv 1 + 4 + 9+6 + 5+ 6+ 9 + 4 +1 + 0= 45\equiv 5\pmod{10}$$

Therefore, the units digit of $1^2 + \cdots + 2010^2$ equals $$4 \times 2010/10 = 5\times 201 \equiv 5\pmod{10}$$

Meanwhile $$2011^2+2012^2+2012^2+2013^2+2014^2+2015^2\equiv 1^2 + 2^2 + 3^2+4^2+5^2\equiv 1+4+9+6+5\equiv 5\pmod{10}$$

Therefore, the final result is $$5+5\equiv \boxed{0}\pmod{10}$$

$\underline{Alternative\ Solution}$

By the sum of square formula, we find

$$1^2 +\cdots + 2015^2 = \Big(\frac{2015\times 2016}{2}\Big)^2=(2015\times 1008)^2\equiv (5\times 8)^2\equiv \boxed{0}\pmod{10}$$