The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}^n$ so that 
\begin{equation}
\mathcal{H}^n(\{ f \neq g \} \cup \{ Df \neq Dg \} ) < \epsilon.
\end{equation}
This can be deduced from the Whitney extension theorem; you can find a proof in Leon Simon's lecture notes on GMT (Theorem 5.3, pp. 32-33).

In general you cannot find a $C^1$ function $g$ that coincides with a Lipschitz $f$ almost everywhere. Here is a counterexample in the lowest dimension, where $m = n = 1$.

Let $C \subset [0,1]$ be a fat Cantor set. This is closed, has empty interior, and measure $\lambda := \mathcal{H}^1(C) \in (0,1)$. Define $f: x \mapsto \int_0^x \mathbf{1}_C$. This is $1$-Lipschitz and has $f(1) = \lambda$; moreover $f' = 0$ on the open complement $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ in $C^c$. As $C$ has empty interior, this means that $g$ is constant, and cannot coincide with $f$ a.e.