This appears not to be the case: there is a $F_\sigma$ set $B$ such that $S_B$ is not Borel. This is optimal since the bullets in the question explain how $S_B$ is Borel when $B$ is $G_\delta$.

There is surely a better way to explain this. I arrived at this answer by first realizing that the answer is yes if we replace "$[p] \subseteq B$" by "$[p]\cap B$ is comeager in $[p]$" in the question. I then attempted to transform this into a positive answer to the question, and my failure to do so led to what follows.

Let $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ be a sequence of nonempty subtrees of $2^{<\omega}$ with no dead ends. For a (possibly empty) tree $p \subseteq 2^{<\omega}$, define $$p' = \{ t \in p \mid (\forall n)(p_t \nsubseteq c_n)\},$$ where $p_t = \{ s \in p \mid s \subseteq t \lor t \subseteq s \}.$

Topologically: $c_0,c_1,c_2,\ldots$ is a sequence of codes for closed sets $[c_0] \subseteq [c_1] \subseteq [c_2] \subseteq \cdots$ If $U_i$ is the relative interior of $[c_i]\cap[p]$ in $[p]$, then $[p'] = [p] \setminus \bigcup_{i<\omega} U_i.$ By the Baire Category Theorem, if $[p] \cap \bigcup_{i<\omega} [c_i]$ is comeager in $[p]$ then $[p] \cap \bigcup_{i<\omega} U_i$ is open dense in $[p]$ and thus $[p']$ is nowhere dense in $[p]$.

So, as a first approximation to determining whether $[p] \subseteq \bigcup_{i<\omega} [c_i]$ we can check that $[p']$ is comeager in $[p]$. This is a $\Pi^0_2$ check: $(\forall t \in p)(\exists u \in p)(t \notin p' \land t \subseteq u).$ This is not enough however as we have merely reduced the question to whether $[p'] \subseteq \bigcup_{i<\omega} [c_i].$

We can iterate this operation: define $p^{(0)} = p,$ $p^{(\alpha+1)} = (p^{(\alpha)})',$ and $p^{(\alpha)} = \bigcap_{\beta<\alpha} p^{(\beta)}$ when $\alpha$ is a limit ordinal. Since $p$ is a countable set, there is always some $\alpha<\omega_1$ such that $p^{(\alpha+1)} = p^{(\alpha)}$ and the sequence stabilizes from then on. Let's call $p^{(\alpha)}$ the *core of $p$* and lets call the first such $\alpha$ the *core rank of $p$*. For convenience, let's write $p^\ast$ for the core of $p$.

Note that $[p] \subseteq \bigcup_{i<\omega} [c_i]$ if and only if the core of $p$ is empty.
The *only if* direction is follows from the observation that we always have $[p] \subseteq [p'] \cup \bigcup_{i<\omega} [c_i]$. For the *if* direction, notice that the core $p^\ast$, when nonempty, has the property that $p^\ast \setminus c_n$ is dense in $p^\ast$ for every $n$. So a generic path through $p^\ast$ witnesses that $[p^\ast] \subseteq [p] \nsubseteq \bigcup_{i<\omega} [c_i].$

For perfect $p$, there are $c_0 \subseteq c_1 \subseteq c_2 \subseteq\cdots$ with empty $p^\ast$ of arbitrarily large core rank with respect to $p$. For simplicity, we take $p = 2^{<\omega}.$ To get started, note that if $$c_i = \{0\}^{<\omega} \cup \{ s \in 2^{<\omega} \mid |s| > i \land (\exists j \leq i, s_j = 1)$$ then $2^\omega = \bigcup_{i<\omega} [c_i]$ and $c_0,c_1,c_2,\ldots$ has core rank $2$ with respect to $2^{<\omega}$.
To get increasingly larger core ranks, given $c_{k,0} \subseteq c_{k,1} \subseteq \cdots$ for $k = 0,1,2,\ldots$ with $2^{\omega} = \bigcup_{i<\omega} [c_{k,i}].$ Define $$c_i = z \cup \bigcup\nolimits_{k<\omega} \{ 0^{k-1}1 t \mid t \in c_{k,i} \}.$$ Then $2^{\omega} = \bigcup_{i<\omega} [c_i]$ and a straghtforward inductive calculation shows that this sequence has core rank at least $\alpha+1$ where $\alpha$ is any ordinal for which there are infinitely many $k$'s where $c_{k,0},c_{k,1},c_{k,2},\ldots$ has core rank at least $\alpha$. In particular, if $c_{k,0},c_{k,1},c_{k,2},\ldots$ has core rank $\alpha_k$ and $\alpha_0 \leq \alpha_1 \leq \cdots$ then $c_0,c_1,c_2,\ldots$ has core rank $\sup_{k<\omega} (\alpha_k+1).$

Now a sequence $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ can be encoded by an $f \in \omega^\omega$ in such a way that $f(n)$ determines all $c_i \cap 2^n$ at once. This is because $c_0 \cap 2^n \subseteq c_1 \cap 2^n \subseteq \cdots,$ so it suffices for $f(n)$ to encode the finitely many subsets of $2^n$ that occur in this sequence along with the first index at which they appear.

Define the universal sequence $d_0 \subseteq d_1 \subseteq d_2 \subseteq \cdots$ as follows: $t \in d_i$ if the longest initial segment of $t$ which is of the form $$1^{n_0}0s_01^{n_1}0s_1\cdots1^{n_{k-1}}0s_{k-1},$$ where each $n_j$ appropriately encodes an infinite nondecreasing sequence of subsets of $2^j$ according to the scheme above, is such that $s_0s_1\cdots s_{k-1}$ belongs to the $i$th set coded by $n_{k-1}.$

Note that if $f$ is the code for $c_0 \subseteq c_1 \subseteq c_2 \subseteq \cdots$ then the perfect set $p$ where $$[p] = \{1^{f(0)}0x_01^{f(1)}0x_11^{f(2)}0x_2\cdots \mid x \in 2^\omega\}$$ is such that $p,d_0 \cap p,d_1 \cap p, d_2 \cap p,\ldots$ is isomorphic to $2^{<\omega},c_0,c_1,c_2,\ldots$ It follows that for this sequence $d_0 \subseteq d_1 \subseteq d_2 \subseteq \cdots$ there are $p$ with $p^\ast = \varnothing$ and arbitrarily large core rank with respect to $d_0,d_1,d_2,\ldots$

Let $B = \bigcup_{i<\omega} [d_i]$, which is $F_\sigma.$ For each $\alpha<\omega_1,$ the set $S_\alpha = \{ p \mid p^{(\alpha)} = \varnothing \}$ is Borel, where $p^{(\alpha)}$ is computed with respect to $d_0,d_1,d_2,\ldots$, and if $\alpha \leq \beta < \omega_1$ then $S_\alpha \subseteq S_\beta$.
We now have $S_B = \bigcup_{\alpha<\omega_1} S_\alpha$ but $S_B \nsubseteq S_\alpha$ for any $\alpha <\omega_1$. Therefore $S_B$ is not Borel.