Another try, now I claim that the inequality holds.

Denote $T:=R^{1/\sigma-1}$, then $T<1$. Let $n=N-n_1$ and $\xi_1,\ldots,\xi_n$ be i.i.d. taking the values 1 or $T$ with probabilities $p$ and $1-p$ respectively. Then, $B(n_1)=(\mathbb{E}(\xi_1+\ldots+\xi_n)^\sigma)^{1/\sigma}$. We use quasilinearization trick: if $1/\sigma+1/\tau=1$, then by $L^\sigma-L^\tau$ duality we have $$ (\mathbb{E}(|X|^\sigma))^{1/\sigma}=\max_{Y\colon \mathbb{E} |Y|^\tau=1} \mathbb{E}(XY), $$ where the maximum is taken over all random variables $Y$ with $\tau$'th moment equal to 1. We use this for $X=\xi_1+\ldots+\xi_n$ and find the appropriate $Y$. Note that $\mathbb{E} |Y|\leqslant 1$ (power means inequality). Then $$B(n_1)=\mathbb E(XY) =\mathbb{E} (\xi_1+\ldots+\xi_{n-1})Y+\mathbb E \xi_n Y\\\leqslant (\mathbb{E} (\xi_1+\ldots+\xi_{n-1})^\sigma)^{1/\sigma}+\mathbb{E} |Y|\leqslant B(n_1+1)+1,$$ where we used Holder inequality, $|\xi_n|\leqslant 1$ and $\mathbb{E} |Y|\leqslant 1$.

Another try, now I claim that the inequality holds.

For a random variable $X$, denote $\|X\|=(\mathbb E |X|^\sigma)^{1/\sigma}$. It is indeed a norm. 

Then $$ \|X_1+\ldots+X_n\|\leqslant \|X_1+\ldots+X_{n-1}\|+\|X_n\|\leqslant \|X_1+\ldots+X_{n-1}\|+1. $$

Without additional assumptions this can not be true. If you fix the value of $T:=R^{1/\sigma-1}$ and let $\sigma$ tend to $1$, $B(n_1)$ tends to $(N-n_1)(p+(1-p)T)$ (as this is the expectation of $\xi_1+\ldots+\xi_{N-n_1}$, where $\xi_i$ are i.i.d. taking the values 1 or $T$ with probabilities $p$ and $1-p$ respectively). Then, since $T>1$, the inequality holds with the opposite sign.

Another try, now I claim that the inequality holds.

Denote $T:=R^{1/\sigma-1}$, then $T<1$. Let $n=N-n_1$ and $\xi_1,\ldots,\xi_n$ be i.i.d. taking the values 1 or $T$ with probabilities $p$ and $1-p$ respectively. Then, $B(n_1)=(\mathbb{E}(\xi_1+\ldots+\xi_n)^\sigma)^{1/\sigma}$. We use quasilinearization trick: if $1/\sigma+1/\tau=1$, then by $L^\sigma-L^\tau$ duality we have $$ (\mathbb{E}(|X|^\sigma))^{1/\sigma}=\max_{Y\colon \mathbb{E} |Y|^\tau=1} \mathbb{E}(XY), $$ where the maximum is taken over all random variables $Y$ with $\tau$'th moment equal to 1. We use this for $X=\xi_1+\ldots+\xi_n$ and find the appropriate $Y$. Note that $\mathbb{E} |Y|\leqslant 1$ (power means inequality). Then $$B(n_1)=\mathbb E(XY) =\mathbb{E} (\xi_1+\ldots+\xi_{n-1})Y+\mathbb E \xi_n Y\\\leqslant (\mathbb{E} (\xi_1+\ldots+\xi_{n-1})^\sigma)^{1/\sigma}+\mathbb{E} |Y|\leqslant B(n_1+1)+1,$$ where we used Holder inequality, $|\xi_n|\leqslant 1$ and $\mathbb{E} |Y|\leqslant 1$.