The answer is no. E.g., let $d=2$ and let $Z_1,Z_2,\dots$ be iid standard normal random variables. Let then $G_i=(a_iZ_i,0)$ if $i$ is odd and $G_i=(0,a_iZ_i)$ if $i$ is even, where the $a_i$'s are positive real numbers increasing fast enough in $i$ so that $\sum_1^{n-1}a_i^2=o(a_n^2)$; the convergence everywhere here is as $n\to\infty$. (For instance, one may take $a_i=i!$ or $a_i=2^{2^i}$.) For any random vector $X$, let $\|X\|_2:=(E\|X\|^2)^{1/2}$. For any sequence $(X_n)$ of random vectors and any sequence $(b_n)$ of positive real numbers, write $X_n=o_2(b_n)$ to mean $\|X_n\|_2=o(b_n)$.
Then $\sum_1^{n-1} a_{2j-1}Z_{2j-1}=o_2(a_{2n-2})$ and $\sum_1^{n-1} a_{2j}Z_{2j}=o_2(a_{2n-1})$.
Let
\begin{equation}
S_n:=\sum_1^n G_i.
\end{equation}
Then for natural $n$ we have
\begin{equation}
S_{2n}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^n a_{2j}Z_{2j}\Big)=a_{2n}(0,Z_{2n})+o_2(a_{2n}),
\end{equation}
\begin{equation}
S_{2n-1}=\Big(\sum_1^n a_{2j-1}Z_{2j-1},\sum_1^{n-1} a_{2j}Z_{2j}\Big)=a_{2n-1}(Z_{2n-1},0)+o_2(a_{2-1}),
\end{equation}
\begin{equation}
\sigma_{2n}^2=ES_{2n}S_{2n}^T=a_{2n}^2
\begin{pmatrix}
o(1)&o(1)\\o(1)&1+o(1)
\end{pmatrix},
\end{equation}
\begin{equation}
\sigma_{2n-1}^2=ES_{2n-1}S_{2n-1}^T=a_{2n-1}^2
\begin{pmatrix}
1+o(1)&o(1)\\o(1)&o(1)
\end{pmatrix},
\end{equation}
\begin{equation}
\|\sigma_n\|^2\sim a_n^2.
\end{equation}
So, the matrix
\begin{equation}
M_n:=\frac1{\|\sigma_n\|^2}E\Big(\sum_1^n G_i\Big)\Big(\sum_1^n G_i\Big)^T=\frac{\sigma_n^2}{\|\sigma_n\|^2}
\end{equation}
does not converge to any matrix; rather, $M_n$ asymptotically oscillates between $\begin{pmatrix}
0&0\\0&1
\end{pmatrix}$ and $\begin{pmatrix}
1&0\\0&0
\end{pmatrix}$.
