Because $3^2\equiv 1\pmod{8}$, we find for any even $k$, $3^k\equiv 1\pmod{0}$. Accordingly, for odd $k$, we have $3^k\equiv 1\times 3\equiv 3\pmod{8}$. It follows that $$3^0 + 3^1 + 3^2 + \cdots + 3^{2009}\equiv 1+3+1+\cdots + 3\equiv \boxed{4} \pmod{8}$$