Let $g(n) = 3(-1)^n\binom{2n}{n}$. The conjecture states $(g * \mu)(n)$ is divisible by $n^3$. We'll show the divisibility prime-power-wise.

Consider any $p|n$. We can couple summands in $g * \mu$ as follows: divisors $d_1,d_2$ of $n$ are coupled iff $d_1=pd_2$ and $\mu(n/d_1), \mu(n/d_2)\neq 0$. We can now rewrite $(g * \mu)(n)$ as a sum over couples: $\sum_{(d_1,d_2)} \mu(n/d_1) (g(d_1) - g(d_2))$. This way we see it is enough to show that $p^{3k} \mid g(dp^k)-g(dp^{k-1})$ for any prime $p$ and $(d,p)=1$.

For $p \ge 5$, this follows directly from the case $\epsilon=0$ of Lemma A in
Gessel's Paper:

Let $p$ be a prime. Let $\epsilon=1$ if $p$ is 2 or 3, and $\epsilon=0$ if $p$ is greater than $3$. Then $\binom{p^ka}{p^k b} \equiv \binom{p^{k-1}a}{p^{k-1}b} \pmod {p^{3k-\epsilon}}$

This means that the conjecture is true for any $n$ not divisible by $2,3$.

For $p=3$, this also follows from the lemma from the $\epsilon=1$ case, thanks to the fact that the multiplicative term 3 appears in $g$.

So it remains to work out $p=2$. From the same lemma, we find that for $p=2$, $p^{3k-1} \mid g(dp^k)-g(dp^{k-1})$, so $a(n)$ is at worst an half-integer.

The proof of the lemma relied on theorem 2.2. Specializing the theorem to the case $p=2, a=2^{k+1}t,b=2^k t$ we find (using the fact that $v_{2}(\binom{2n}{n}) \ge 1$):

If $t$ is odd and $k\ge 2$, $\binom{2^{k+1}t}{2^{k}t} \equiv \binom{2^k t}{2^{k-1}t} \pmod {2^{3k}}$. If $t$ is odd and $k=1$, $\binom{2^{k+1}t}{2^{k}t} \equiv -\binom{2^k t}{2^{k-1}t} \pmod {2^{3k}}$

So $g(2^k t) \equiv g(2^{k-1}t) \pmod {2^{3k}}$ (only in the case $k=1$ we used the $(-1)^n$ term in $g(n)$). Hence the conjecture is proved.