Van Hamme's supercongruence (A.2) for a truncated hypergeometric series

Let $p$ be an odd prime. For parameters $a_1,,a_6$ and $b_1,,b_5$, write the truncated hypergeometric series as

$${}_{6}F_{5}\left[ \begin{matrix} a_1,a_2,a_3,a_4,a_5,a_6\\ b_1,b_2,b_3,b_4,b_5\end{matrix};z\right]_n=\sum_{k=0}^n\frac{(a_1)_k(a_2)_k(a_3)_k(a_4)_k(a_5)_k(a_6)_k}{(b_1)_k(b_2)_k(b_3)_k(b_4)_k(b_5)_k}\frac{z^k}{k!},$$

where $(a)_0=1$ and $(a)_k=a(a+1)\cdots(a+k-1)$ for $k\geq 1$. Let $\Gamma_p(\cdot)$ denote the $p$-adic Gamma function. Van Hamme's supercongruence (A.2).

$${}_{6}F_{5}\left[ \begin{matrix} \frac{5}{4},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\\ \frac{1}{4},1,1,1,1\end{matrix};-1\right]_{\frac{p-1}{2}}\equiv\begin{cases}-p\Gamma_p\left(\frac{1}{4}\right)^4\pmod{p^3}&\text{if }p\equiv1\pmod{4},\\0\pmod{p^3}&\text{if }p\equiv3\pmod{4}. \end{cases}$$

The congruence was first confirmed by McCarthy and Osburn, so it is solved rather than open.