Let $g$ be a normal order of an arithmetic function $f$, i.e. for every $\epsilon > 0$, $$\left| \frac{f(n) - g(n)}{g(n)} \right| < \epsilon$$ holds for almost all $n \in \mathbb{N}$. Does this imply that $g$ is an average order of $f$, i.e. $$\sum_{n \le x} f(n) \sim \sum_{n \le x} g(n)$$ as $x \rightarrow \infty$? I think this should be true when $| \sum_{n \le x} g(n)| \rightarrow \infty$ but it might not be true in general. Can you prove/disprove the above claim? Also, the converse is not true as well.
Edit: $\omega(n)$, the no. of distinct prime divisors of $n$, is a good counterexample for the converse because it has average order $\log \log n$ but it cannot have normal order $\log \log n$ as even for large $n$, $\omega(n)$ could be as small as $1$.
Edit 2: The above paragraph is incorrect, as specified in the accepted answer.
First, your last paragraph is incorrect: $\omega(n)$ does indeed have normal order $\log\log n$ (this is a theorem of Hardy—Ramanujan). The fact that $\omega(n)=1$ for infinitely many $n$ does not contradict this normal-order statement, since the set of $n$ for which $\omega(n)=1$ has density $0$.
The answer to your main question is no. Define for example $g(n)=1$ everywhere, and $f(n)=n$ if $n$ is prime and $f(n)=1$ if $n$ is not prime. Then $f(n)$ has normal order $1=g(n)$, since the primes have density $0$. However, $$ \sum_{n\le x} f(n) \sim \frac{x^2}{\log x} \quad\text{while}\quad \sum_{n\le x} g(n) \sim x. $$ In general, a function can have all kinds of arbitrary behavior on a set of density $0$ yet still have a normal order.