Answer: 46

Solution:

The sum of the logarithms is $\log_8 (a_1 a_2 \cdots a_{12}) = \log_8\!\left(a^{12} r^{66}\right),$ so $a^{12} r^{66} = 8^{2006} = 2^{6018},$ which gives $a^2 r^{11} = 2^{1003}.$

Thus $a$ and $r$ are powers of $2:$ write $a = 2^x$ and $r = 2^y$ with integers $x, y \ge 0$ and $2x + 11y = 1003.$ Since $2x$ is even and $1003$ is odd, $y$ must be odd, say $y = 2k - 1$ for $k \ge 1.$ Then $x = 507 - 11k \ge 0$ exactly when $k \le \lfloor 507/11 \rfloor = 46.$

Each $k = 1, 2, \ldots, 46$ gives one pair, so there are $46$ ordered pairs $(a, r).$