$\newcommand{\norm}[1]{\Vert#1\Vert}$

In general, I think the answer to your question is no. Take ${\mathcal B}=C^1[-1,1]$ with the norm
$\norm{f}= \norm{f}_\infty+\norm{f'}_\infty$
and let ${\mathcal I}$ be the closed ideal consisting of those $C^1$-functions which vanish at $x=0$ and whose 1st derivative vanishes at $x=0$. Then $I\cap {\mathcal B}$ contains the function $f(x)=x$ which is evidently not in ${\mathcal I}$.

[Some general remarks follow, in a rambling style owing to lack of sleep. I may try to edit these later.]

In the commutative unital setting, taking $B=C(X)$, we know what the closed ideals of $B$ are (they are precisely the "kernels" of closed subsets of $X$, in the language of hulls and kernels).

If your subalgebra ${\mathcal B}$ also has maximal ideal space (homeo to) $X$, then your question is related to -- perhaps is equivalent to, I have not thought in detail -- the following one:

**Can I find a closed two sided ideal in ${\mathcal B}$ which is not the kernel of its hull?**

Without your restrictions on stability-under-func-calc, this kind of question has been much studied for commutative examples, and I think also for certain noncommutative examples related to group algebras.

For little Lipschitz algebras (on the circle) the answer is no -- this ought to be in a paper of Sherbert from the 1970s
-- so I expect the answer to your original question is "yes". (For the "big" Lipschitz algebras my suspicion is that the counter-example I gave for $C^1[-1,1]$ would also work.)