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 $\mathrm{SO}(4)$.
I don't have an example to hand, but I suspect that the answer to your second question is also 'no'. I'll try to write down an example when I have time.