Actually, the contours $C$ in the two integrals are different.
By definition, $C$ goes from $+\infty$ through an anticlockwise path return to $+\infty$, encircling all poles of $\prod_{j=1}^m\Gamma(b_j-s)$ (each pole exactly once) and none pole of $\prod_{j=1}^n\Gamma(1-a_j-s)$. In this case, the integral-contours of $\textrm{G}^{4, 0}_{2, 4}\Big({1,c+\frac12\atop c,c,c,1}\Big|1\Big)$ encircles poles:
$$c, c+1, c+2,\dotsc, c+k,\dotsc,$$
and poles:
$$0, -1, -2, \dotsc, k, \dotsc,$$
must be outside the contours. Similar for $\textrm{G}^{3, 1}_{2, 4}\Big({1,c+\frac12\atop c,c,c,1}\Big|1\Big)$, we needs the contour encircles poles:
$$0, 1, 2, \dotsc, k,\dotsc,\quad \mathrm{and}\quad c, c+1,c+2,\dotsc,c+k,\dotsc,$$
As you have found, $1,2,\dotsc,n,\dotsc$ are actually normal points of the integrand, so the contour can "deform" over these points. However, $0$ remains to be a pole, so the contour can not cross over it, which makes the expression obtain a non-zero value.
Thus, for any $c\neq0,-1,-2,\dotsc,-k,\dotsc$ (by definition, they're not allowed), the sum of this two Meijer functions is:
$$-\operatorname{Res}\frac{[\Gamma(c-s)]^3}{s\,\Gamma\big(c+\frac12-s\big)}\Bigg|_{s=0} =-\dfrac{[\Gamma(c)]^3}{\Gamma\big(\frac12+c\big)}.$$
Using Mathematica, we can check our answer:
Table[-MeijerG[{{},{1,c+1/2}},{{0,c,c,c},{}},1,-1]
-MeijerG[{{1},{c+1/2}},{{c,c,c},{0}},1,-1]
+Gamma[c]^3/Gamma[1/2+c]//N,
{c, SetPrecision[RandomReal[{1,2}, 10]
+I RandomReal[{1, 2}, 10], 10]}
]//Column
Notice that in Mathematica
-MeijerG[{{__}, {__}}, {{__}, {__}}, _, -1]
is corresponding to the definition in Wikipedia(Bateman & Erdelyi).