One of the best-known classes is the "quasi-polynomials", which are exponentials of polynomials in logs, e.g. $e^{\log^2(x)+\log x}$, which you might also write as $x^{\log(x)+1}$. As long as the degree of the exponent is greater than $1$, these fit between polynomial and exponential.
One has also the "sub-exponentials," which grow as $e^\phi$ where $\lim\limits_{x\to \infty}\frac{\phi(x)}{x}=0$. The most obvious examples that aren't quasi-polynomial are along the lines of the one you gave.
These don't exhaust the possibilities, though. You may be interested in a considerable volume of discussion over at MO of functions $f$ such that $f(f(x))$ is exponential.
These "half-exponentials" are in between the two classes I've described: a proper sub-exponential has an exponent that dominates all polynomials of logarithms, so its composition with itself has an exponent that dominates all polynomials-and thus isn't exponential. In the other direction, you can see that quasi-polynomials are closed under self-composition. Here's the latest thread, with links to others.
https://mathoverflow.net/questions/45477/closed-form-functions-with-half-exponential-growth
Here is a more general framework for your question. Suppose you have functions $f(x)$ and $g(x)$. You compare their growth rates by looking at $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$
In your problem, $f(x)=3^x$ and $g(x)=2^x$, and this limit is $\infty$, which is what we mean when we say "$3^x$ grows faster than $2^x$."
Then you ask: why does taking the logarithm not yield the same order? (In other words, you're asking: if $f(x)$ grows faster than $g(x)$, why doesn't $\log f(x)$ grow faster than $\log g(x)$? Or, in yet other words: why don't logs preserve the asymptotic ordering?) The answer is that you must now consider $$\lim_{x\to\infty}\frac{\log f(x)}{\log g(x)}$$
And there is no nice way to relate the first limit to this limit, in general.
But in your special case when $f(x)$ and $g(x)$ are exponentials, taking the log gives first-order polynomials, which have the same growth rate: $$\lim_{x\to\infty}\frac{\log 3^x}{\log 2^x}=\lim_{x\to\infty}\frac{x\log 3}{x\log 2}=\frac{\log 3}{\log 2}$$
(Since the limit is finite and non-zero, the new functions have the same growth rate.)
EDIT
No, pure exponentials are not the only family of functions that has this property. Consider $f(x)=x\cdot 3^x$ and $g(x)=x\cdot 2^x$. These functions have different growth rates, because $$\lim_{x\to\infty}\frac{x\cdot 3^x}{x\cdot 2^x}=\infty$$
But their logs, $\log x+x\log 3$ and $\log x+x\log 2$, have the same growth rates, because
$$\lim_{x\to\infty}\frac{\log x+x\log 3}{\log x+x\log 2}=\lim_{x\to\infty}\frac{\frac{1}{x}+\log 3}{\frac{1}{x}+\log 2}=\frac{\log 3}{\log 2}$$
In fact, as this example suggests, any function that is the product of a bunch of functions all of which have growth rate at at most exponential will work: their logarithms will have the same growth rate. Take $f(x)=(\log_3 x)(x^3-4x^2+1)3^x$ and $g(x)=(\log x)(x^2)2^x$ for example. These functions have different growth rates, again, but their logarithms have the same growth rate. The logarithm turns the multiplicative structure into an additive structure dominated by a first-order polynomial.
Best Answer
Here's a sketch of a proof: suppose $a,b>0$, and $a\neq b$. Without loss of generality, $a>b$. We want to show that $O(a^n)\neq O(b^n)$, or equivalently, that $a^n\notin O(b^n)$ (why are these equivalent?)
To show that $a^n\notin O(b^n)$, it suffices to show (again, why?) that
$$ \lim_{n\rightarrow\infty}\frac{a^n}{b^n}=\infty $$ Using the fact that $a>b>0$, this limit should be easy to show.