the answer for your question is yes. Note that, all you need to prove is the following equality $$ \int_{E} X\ dP = \int_{E} \left[\int_{\Omega} X\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} \qquad \qquad (1). $$ for all $E\in\mathcal{N}$.
The simplest way to do this is start from characteristic functions and then use linearity and finally the powerfull approximation theorems of the measure theory.
Let us start with the case $X=1_{A}$ where $A\in \mathcal{F}$. In this case the above equality holds as long as $$ P(A\cap E)=\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}}. $$ But this is certainly true and we can prove this manipulating the rhs. Indeed $$ \int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} = \int_{\Omega} \mathbb{E}(A\cap E|\mathcal{N}) dP|_{\mathcal{N}} = \int_{E} \mathbb{E}(A|\mathcal{N}) dP|_{\mathcal{N}}, $$ where we used in the last equality that $E$ is $\mathcal{N}$ measurable. But the right hand side above is by the properties of conditional expectation equals to $P(A\cap E)$. So the identity (1) holds for any characteristic function and for all $E\in\mathcal{N}$.
By linearity of the integral (1) is certainly true if $X$ is any simple function. Let $X$ be any positive $\mathcal{F}-$measurable function. We know that there is a sequence of simple functions $\varphi_n \uparrow X$, here the up arrow means monotone convergence. Since at this point is clear for us that $$ \begin{eqnarray} \int_{E} \varphi_n\ dP &=& \int_{E} \left[\int_{\Omega} \varphi_n\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} \\ &=& \int_{E} \mathbb{E}(\varphi_n|\mathcal{N}) dP|_{\mathcal{N}}, \end{eqnarray} $$ we can finish the proof by invoking the Monotone Convergence Theorem.
Edition. I corrected the error pointed out by Didier Piau.
Addition. Theorem. Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measure space, $\mathcal{A}$ a sub-$\sigma$-algebra of $\mathcal{B}$, and $\nu=\mu|_{\mathcal{A}}$. If $f\in L^1(\mu)$, there exist $g\in L^1(\nu)$ (thus $g$ is $\mathcal{A}$-measurable) such that $$ \int_{E} f\ d\mu =\int_{E} g\ d\nu $$ for all $E\in A$; if $g'$ is another such function then $g=g'$ $\nu$-a.e. (Note that $g$ is the conditional expectation of $f$ on $\mathcal{A}$. (Folland)
