Let $L$ be an embedded real $n$-dimensional manifold in $\mathbb C^n$, where $n\ge 2$.
Given a point $p \in L$, does there exists a neighborhood $U$ of $p$ and holomorphic function $$\phi:U \rightarrow \mathbb C^n,$$ such $\phi$ is a diffeomoprhism onto its image and $$\phi(U \cap L) \subset \mathbb R^n \subset \mathbb C^n = \mathbb R^n \oplus i \mathbb R^n?$$
My guess is this is not true in general. I would love to know why not and also what extra conditions, if any, would make it true.
On
As you correctly guessed, the answer to your question is "no" in general.
For example in $\mathbb C^2$ with coordinates $z,w$ the real $2$- dimensional submanifold $L=\{w=0\}\subset \mathbb C^2$, which happens to be also a complex submanifold, will be transported by any holomorphic isomorphism $\mathbb C^2 \stackrel \simeq \to \mathbb C^2$ to a holomorphic submanifold $\phi(L)\subset \mathbb C^2$. Hence we will never have $\phi(L)\subset\mathbb R^2$.
(For ease of notations I have considered a global isomorphism $\phi$, but the argument works locally).
There's a fairly straightforward necessary and sufficient condition. First, a couple of definitions:
A $k$-dimensional real submanifold $L\subset \mathbb C^n$ is said to be real analytic if for each $p\in L$ there is a neighborhood $U$ of $p$ and a real-analytic submersion $\phi\colon U\to \mathbb R^{2n-k}$ such that $L\cap U = \phi^{-1}(0)$.
Such a submanifold is said to be totally real if for each $p\in L$, the real-linear map $J: \mathbb C^n\to \mathbb C^n$ obtained by multiplying by $i$ in complex coordinates satisfies $T_pL \cap J(T_pL) = \{0\}$. (In other words, there is no vector $v\in T_pL$ such that $iv$ is also tangent to $L$.)
Theorem. Let $L$ be an embedded real $n$-dimensional submanifold in $\mathbb C^n$, where $n\ge 2$. The following conditions are equivalent.
Sketch of proof: Let's use coordinates $z^j = x^j + i y^j$ on $\mathbb C^n$. For necessity, note that both the real analytic and totally real conditions are satisfied by the submanifold $y^1=\dots = y^n = 0$, and both conditions are preserved by biholomorphic maps.
For sufficiency, suppose $L$ satisfies condition $2$, and let $p\in L$ be given. After applying a translation, we can assume $p$ is the origin. Let $(v_1,\dots,v_n)$ be a basis for the (real) tangent space $T_pL$. The fact that $T_pL$ is totally real ensures that the complex-linear map $A(z^1,\dots,z^n) = z^1v_1 + \dots z^n v_n$ has trivial kernel, and $A^{-1}$ takes $T_pL$ to the subspace defined by $y^1=\dots y^n=0$. Thus after applying $A^{-1}$, we may assume that $T_pL$ is that subspace. Now the real-analytic version of the implicit function theorem shows that there are neighborhoods $U_0,V_0$ of $0$ in $\mathbb R^n$ and a real-analytic map $f\colon U_0\to V_0$ such that $L\cap (U_0\oplus i V_0)$ is the set determined by the equations $y^j = f^j(x^1,\dots,x^n)$. The fact that $T_0L$ is the $y=0$ plane means that all first partial derivatives of each $f_j$ vanish at the origin. By real-analyticity, there is some neighborhood $U_1$ of $U_0$ in $\mathbb C^n$ to which the $f^j$'s extend as holomorphic functions, still denoted by $f^j$.
Define a map $F\colon U_1\to \mathbb C^n$ by $F^j(z) = z^j + if^j(z)$. The fact that the partial derivatives of $f^j$ vanish at the origin means that $dF_0$ is the identity map, so the holomorphic version of the inverse function theorem shows that $F$ is a biholomorphism from some smaller neighborhood of the origin to its image. You can check that $F$ takes a neighborhood of the origin in the set $\{y=0\}$ to a neighborhood of the origin in $L$. $\square$