Your first question is a bit ambiguous. Are you asking whether, for each $p\in U$, the matrices $S_k(p)$ span the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$ or are you asking whether, after taking the span of the union of the images of $S_k(p)$ over all $p\in U$, the result is the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$?
The answer to the first of these two questions is clearly 'no' because you could easily have a point $p$ where all of the $S_k(p)$ vanish and yet the Lie group $\mathrm{Hol}^0_p(\nabla)$ has positive dimension.
Perhaps more surprising, the answer to second version of the question is 'no' as well. One can construct examples of metrics on $\mathbb{R}^4=U=M$ together with a framing $\sigma$ as you describe such that the resulting matrices $S_k$ all take values in $\mathfrak{u}(2)\subset\mathfrak{so}(4)$, but the holonomy of the metric is equal to $\mathrm{SO}(4)$.
The answer to your second question is also 'no'. Here's an example: Let $M=\mathbb{R}^2$ (i.e., the $xy$-plane) and let the vector bundle be the trivial bundle $V = M\times\mathbb{R}^2$. Let $\nabla$ be the connection on $V$ whose connection matrix $\omega$ with respect to the standard basis $\sigma = (\sigma_1,\sigma_2$) of section is $$ \omega = \begin{pmatrix}\mathrm{d}x & \mathrm{d}y\\ \mathrm{d}y & -\mathrm{d}x\end{pmatrix}. $$ Then the corresponding curvature matrix is $$ \Omega = \mathrm{d}\omega + \omega\wedge\omega = \begin{pmatrix}0& 2\,\mathrm{d}x\wedge\mathrm{d}y\\ -2\,\mathrm{d}x\wedge\mathrm{d}y & 0\end{pmatrix}, $$ so it takes values in $\mathfrak{so}(2)$. However, I claim that the holonomy of $\nabla$ is $\mathrm{SL}(2,\mathbb{R})$. To see this, note that the Lie algebra of the holonomy group must contain $\mathfrak{so}(2)$, and, since there is no Lie algebra properly between $\mathfrak{so}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$, it suffices to show that the holonomy of $\nabla$ is not $\mathrm{SO}(2)$.