Conjecture: A binomial sum congruence for Fibonacci

In this related question, @Gerry Myerson asked if the conjecture fails for composites. Upon examining for composites, I found a pattern for module $n^2$ instead of $n^3$ in the linked original question. I have verified these conjectures for $n \le 5 \times 10^6$. Can they be proved or disproved.

Conjecture 1: Let $n$ be a squarefree composite integer satisfying $ n\equiv 1 \text{ or } 19 \pmod{30}. $ If every prime divisor $q$ of $n$ satisfies $$ \left(\frac{-15}{q}\right)=-1, $$ equivalently, $ q\equiv 7,11,13,\text{ or }29\pmod{30}, $ then $$ \sum_{k=0}^{n-1} \binom{2k}{k}^{3} \frac{F_{8k}}{4096^k} \equiv 0\pmod{n^2}. $$ where $F_m$ denotes the $m$th Fibonacci number.

Conjecture 2: Let $n=qr$, where $q$ and $r$ are primes, possibly equal, and suppose that $$ n\equiv 1 \text{ or }19\pmod{30}. $$ Then $$ \sum_{k=0}^{n-1} \binom{2k}{k}^{3} \frac{F_{8k}}{4096^k} \equiv0\pmod{n^2} $$ if and only if $$ \left(\frac{-15}{q}\right) = \left(\frac{-15}{r}\right) = -1. $$ Equivalently, the congruence holds if and only if $ q,r\equiv7,11,13,\text{ or }29\pmod{30}. $ where $F_m$ denotes the $m$th Fibonacci number.