For the sake of simplicity, let us consider only sequences of zeroes and ones.
Notice that if a sequence contains arbitrarily long segments consisting of consecutive ones and arbitrarily long segments of consecutive zeroes, then it is not almost convergent. This can be seen from the fact that for any fixed $p$ you get both ones and zeroes as the value of the fraction
Now it suffices to find a set $A$ such that it has zero density, it contains arbitrarily large segments of consecutive numbers. (The assumption that $A$ has zero density means also that there are arbitrarily large gaps.)
If you choose
1 & n\in A, \\
0 & n\notin A,
then this sequence is statistically convergent to zero, but it is not almost convergent.
One possible way of looking at this is the following. If a set $A$ determines a sequence $(x_n)$ in the way described above, then it converges statistically to zero if and only if $d(A)=0$, i.e., the asymptotic density is zero.
Such sequence is almost convergent to $L$ if and only if $u(A)=L$, where $u(A)$ denotes Banach density (uniform density). I have collected some references about Banach density and it's relation to asymptotic density in this answer: Density of a set of natural numbers whose differences are not bounded.
So you can look at this problem as searching for a set such that $d(A)=0$ but it does not have Banach density.
Basically the same examples are suggested in the paper H. Miller, C. Orhan: On almost convergent and statistically convergent subsequences; DOI: 10.1023/A:1013877718406, Zbl: 0989.40002, MR1924673
I will reproduce here the relevant part.
Proposition 1.1. Almost convergent and statistical convergence are incompatible; i.e., $\mathbf F\nsubseteqq \mathbf S$ and $\mathbf S\nsubseteq\mathbf F$.
Proof. The sequence $s=(s_n)$ defined by $s_n=1$ if $n$ is even and $s_n=0$ if $n$ is odd is almost convergent to $1/2$, but it is not statistically convergent. Now consider the sequence of $0$'s and $1$'s defined as follows
where the blocks of $0$'s are increasing by factors of $100$ and blocks of $1$'s are increasing by factors of $10$. This sequence is not almost convergent but is statistically convergent to zero, which completes the proof.