Yes, you could instead (uniquely!) write $\,p = 2^j a\,$ and $\,q = 2^k b\,$ for $\,a,b\,$ odd and then deduce a contradiction by comparing the number of factors of $2,\,$ viz. $\,p^2\,$ has an even number of $2$'s but $\,2q^2\,$ has an odd number of $2$'s, a contradiction. This method uses only the very easily-proved existence and uniqueness of $2$-factorizations, i.e. representations of the form $\,2^j n,\,$ with $\,n\,$ odd (versus the much more powerful, and much more difficult to prove Fundamental Theorem of Arithmetic = existence and uniqueness of arbitrary prime factorizations).
As for your second question, yes, every fraction $\,A/B\,$ can be written with coprime numerator and denominator $\,a/b\,$ simply by cancelling their gcd $\,c,\,$ i.e. $\,A/B = ca/cb = a/b.\,$ By the maximality ("greatest") property of the gcd it follows that $\,d =\gcd(a,b) = 1,\,$ else $\,cd\,$ would be a common divisor of $\,A,B\,$ larger than the greatest common divisor $\,c=\gcd(A,B),\,$ contradiction. However, as above, it suffices to cancel only common factors of $\,2,\,$ so no knowledge of gcds is required.
What is a rational number? The definition I use is that it's a number that can be written as a ratio of two integers. If your book doesn't say explicitly that the integers can be coprime, it's because the notion of a reduced fraction is fundamental to working with fractions. You're not allowed to work with fractions until you believe that they can be written in a unique reduced form, and that your numerator and denominator will be coprime.
If you wanted to, you can prove this with the fundamental theorem of arithmetic. Suppose your rational number $q$ can be written as the ratio of integers $\frac{a}{b}$. Then by the fundamental theorem of arithmetic, $a$ and $b$ have unique factorizations into primes. So factorize them, and the factors that appear in both will cancel. Then you have a coprime numerator and denominator.
In the proof that you're trying, I don't see the need for contradiction. Statement X and Y imply each other. If you have a ratio of non-coprime numbers, reduce it. If you have a ratio of coprime numbers, multiply them both by 2 and now you have a ratio of non-coprime numbers. (The statements you have written are the contrapositive of this.)
Best Answer
For contradiction, assume $\log_{12} 18$ is rational. Also note it's positive. Then exist positive integers $m,n$ such that $\log_{12}18=\frac{m}{n}$. By definition of $\log$, we get $12^{\frac{m}{n}}=18$. Raise both sides to the $n$'th power: $12^m=18^n$, i.e. $2^{2m}\cdot 3^m=2^n\cdot 3^{2n}$, so by Fundamental Theorem of Arithmetic, $2m=n$ and $m=2n$. But the two equations imply $4n=n$, i.e. $4=1$, contradiction.