Let us factor $30^{2010} = 2^{2010} \cdot 3^{2010} \cdot 5^{2010}$.

Any divisor $d$ of $30^{2010}$ can be written as $d = 2^a 3^b 5^c$ where $0 \leq a, b, c \leq 2010$.

The product of two divisors $d_1 = 2^{a_1} 3^{b_1} 5^{c_1}$ and $d_2 = 2^{a_2} 3^{b_2} 5^{c_2}$ is a perfect square if and only if $a_1 + a_2$, $b_1 + b_2$, and $c_1 + c_2$ are all even.

This is equivalent to $a_1 \equiv a_2 \pmod{2}$, $b_1 \equiv b_2 \pmod{2}$, $c_1 \equiv c_2 \pmod{2}$.

Thus, for each divisor, consider the triple $(a \bmod 2, b \bmod 2, c \bmod 2)$. There are $2 \times 2 \times 2 = 8$ possible such triples.

If we select 9 divisors, by the pigeonhole principle, at least two of them must have the same triple. For these two divisors, the exponents of $2$, $3$, and $5$ have the same parity, so their sum is even for each prime, and thus their product is a perfect square.

Therefore, among any nine divisors of $30^{2010}$, there are two whose product is a perfect square.