Now that the question has been clarified, I can answer it. The answer is 'no', there is no CR-isomorphism $\phi: \mathrm{Heis}\to U$ that is unitary on the holomorphic tangent bundles.
To see this, it's probably better to look at the dual $1$-forms. Let $\alpha:T S^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $HS^{2n{+}1}\subset TS^{2n+1}$ and such that the $2$-form $\mathrm{d}\alpha$ restricts to $HS^{2n{+}1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $HS^{2n{+}1}$. Similarly, let $\beta: T \mathrm{Heis}^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $H\mathrm{Heis}^{2n+1}\subset T\mathrm{Heis}^{2n+1}$ and such that the $2$-form $\mathrm{d}\beta$ restricts to $H\mathrm{Heis}^{2n+1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $H\mathrm{Heis}^{2n+1}$.
If $\phi:\mathrm{Heis}\to U$ were a CR-isomorphism that was unitary on the corresponding holomorphic tangent bundles, one would necessarily have $\phi^*\alpha = \beta$. However, because the integral of $\alpha\wedge(\mathrm{d}\alpha)^n$ over $S^{2n+1}$ is finite, its integral over $U$ is also finite. On the other hand, the integral of $\beta\wedge(\mathrm{d}\beta)^n$ over $\mathrm{Heis}$ is infinite because it is the integral of a left-invariant volume form over a noncompact Lie group. Thus, $\phi$ cannot exist.
By the way, it would not be hard to show, using the method of equivalence, that the inequivalence holds even locally, i.e., there is no open subset of $\mathrm{Heis}^{2n+1}$ that is unitarily CR-diffeomorphic to any open subset of $S^{2n+1}$. However, that would take a little more argument, and I am not sure that the OP would be interested.