Is $G$ a secure PRG?
To give this question its deserved answer, I’ll repeat what Ricky Demer noted in his comment:

$$G(x \oplus 1^s) = F(k,x \oplus 1^s) \oplus F(k,x \oplus 1^s \oplus 1^s)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x \oplus 1^s \oplus 1^s) = F(k,x \oplus 1^s) \oplus F(k,x \oplus 0^s)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x \oplus 0^s) = F(k,x \oplus 1^s) \oplus F(k,x)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x) = F(k,x) \oplus F(k,x \oplus 1^s)  \\
\downarrow  \\
F(k,x) \oplus F(k,x \oplus 1^s) = G(x)$$

As tylo commented: This shows G is not a secure PRG.