If I understood your question correctly, I think that the answer is no. In fact, even more is true: if you choose any identification $\varphi \colon \Sigma_1 \to \Sigma_2$ in such a way that the boundary are compatibly identified, then for any embedding of $\Sigma_1$ and $\Sigma_2$ there is a point $p$ of $\Sigma_1$ such that the tangent plane to $\Sigma_1$ is the same as the tangent plane to $\Sigma_2$ at $\varphi(p)$. To see this, let $\mathbb{R}P^2$ denote the real projective plane, the space parameterizing linear subspaces of dimension one in $\mathbb{R}^3$. Suppose by contradiction that you found embeddings with the mentioned property. Let $\gamma \colon \Sigma_1 \cup \Sigma_2 \to \mathbb{R}P^2$ be the function defined by sending $p$ to the linear subspace spanned by $N_p \wedge N_{\varphi(p)}$, where $N_p$ is the normal direction to the image of $p$ and similarly for $N_{\varphi(p)}$, and the wedge product is the usual wedge product in $\mathbb{R}^3$. The function $\varphi$ is indeed a function, since never the normal directions at $p$ and $\varphi(p)$ are parallel. But observe that $\gamma(p)$ is always a vector contained in the tangent plane at $p$. Thus, $\gamma$ determines a global vector field on the image of $\Sigma_1 \cup \Sigma_2$ that is never zero. This is obviously impossible!

Note that the above argument is essentially the one used in the Borsuk-Ulam Theorem (http://en.wikipedia.org/wiki/Borsuk–Ulam_theorem).