Let $f : \mathbb{R}^n \to \mathbb{R}^k$ and let $\nabla$ be the gradient operator.
Under, what condition can we interchange \begin{align} \nabla_u E[f(X+u)]= E[ \nabla f(X+u)] \end{align}
I am well aware of the condition for the one dimensional case. See for example https://en.wikipedia.org/wiki/Leibniz_integral_rule
By analogy with the 1 dimensional case, I think what you really meant to ask is whether something like the following is true. $$ \nabla_t E[f(X,t)]\overset{?}{=}E[\nabla_t f(X,t)] $$ In effect we are passing the derivative with respect to $t$ through the expectation with respect to $x$.
This interchange of differentiation and integration is valid under the exact same conditions as for the one dimensional case. Indeed, such an equality is equivalent to the 1 dimensional interchange holding for each component of the vector $t=(t_1,\ldots,t_n)$, that is, $$ \frac{\partial E[f(X,t]}{\partial t_i}=E\frac{\partial f}{\partial t_i}(X,t),\qquad i=1,\ldots,n. $$