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
$$
\omega = \begin{pmatrix}\mathrm{d}x & \mathrm{d}y\\ \mathrm{d}y & -\mathrm{d}x\end{pmatrix}.\tag0
$$
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 $\mathfrak{so}(2)$. 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 $\mathrm{SO}(2)$.
Suppose that it were. Then there would exist a mapping $g:M\to \mathrm{GL}(2,\mathbb{R})$ and a $1$-form $\alpha$ such that
$$
\omega = g^{-1}\,\mathrm{d}g + g^{-1}\begin{pmatrix}0&\alpha\\-\alpha&0\end{pmatrix}g.\tag1
$$
However, this would imply, by the standard calculation, that
$$
\Omega = g^{-1}\begin{pmatrix}0&\mathrm{d}\alpha\\-\mathrm{d}\alpha&0\end{pmatrix}g.
$$
Consequently, we would have to have
$$
g^{-1}\begin{pmatrix}0&1\\-1&0\end{pmatrix}g = \begin{pmatrix}0&f\\-f&0\end{pmatrix}
$$
for some function $f$ and, taking determinants, we find that $f^2 = 1$, so $f\equiv\pm 1$. This then implies, by algebra, that $g$ is of the form
$$
g = \begin{pmatrix}u&v\\\mp v&\pm u\end{pmatrix}
$$
for some functions $u$ and $v$ on $M$ that do not simultaneously vanish. Plugging this into the supposed formula (1) for $\omega$, we get that $\omega$ must be of the form
$$
\omega = \begin{pmatrix}\mu &\nu\\-\nu&\mu\end{pmatrix}
$$
for some $1$-forms $\mu$ and $\nu$. But this obviously contradicts the definition (0) of $\omega$.
Hence, the holonomy of $\nabla$, which is a connected Lie group properly containing $\mathrm{SO}(2)$ and contained in $\mathrm{SL}(2,\mathbb{R})$, must be equal to $\mathrm{SL}(2,\mathbb{R})$.
Added example: Let $M=\mathbb{R}^3$ with standard coordinates $(x^1,x^2,x^3)$, let the vector bundle be $V = TM$, and let $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ where $\sigma_i$ is the $i$th-coordinate vector field. Let $\nabla$ be the connection that satisifies $\nabla_{\sigma_i}\sigma_i = 0$ and $\nabla_{\sigma_j}\sigma_i = \sigma_k$ whenever $i$, $j$, and $k$ are distinct. Then $\nabla$ is a torsion-free connection on $V=TM$ and the curvature matrix $\Omega$ is skew-symmetric, with the $S_k(p)$ spanning $\mathfrak{so}(3)$ at every point $p\in M$. Nevertheless, the holonomy of $\nabla$ is $\mathrm{SL}(3,\mathbb{R})$, not $\mathrm{SO}(3)$. This follows from an argument very similar to the one given above above.
