I think the answer is yes. Take a basis for $U$ and form a matrix where these are
the rows. We can then put the matrix in row-reduced echelon form and call it $A$.
Each column then has at most one 1. By permuting the columns, you can further suppose that
each row consists of a block of 1's; the lengths of the blocks are non-decreasing and all the 0's are at the end. Call the resulting matrix $\tilde A$, and let the subspace it spans be $\tilde U$. Putting in row-reduced echelon form does not change the subspace spanned, so that the rows of $A$ still span $U$. Applying the column permutation does not change
Hamming weights of vectors, so that $\tilde U$ has the same distribution of Hamming weights as $U$.

Assume that $k=0.99n$, so that each $\tilde A$ has $0.99n$ rows. At most $0.02n$ of these can have two 1's.

Let $W(x)$ denote the Hamming weight of the vector $x$. Your question is whether
$\mathbb P_{\tilde U}(W(x)=\ell)\le C\mathbb P_{\mathbb F_2^n}(W(x)=\ell)$ (where $\mathbb P_V$ is the uniform distribution on the subspace $V$).

Let the rows of $\tilde A$ be $v_1,\ldots,v_{k}$. Any vector in $\tilde U$ can be uniquely expressed as $\epsilon_1 v_1+\ldots+\epsilon_k v_k$. Its weight is
$\epsilon_1 W(v_1)+\ldots+\epsilon_k W(v_k)$. Of course $\epsilon_k$ is uniformly distributed on $\{0,1\}$. Given this we are asking about the distribution of the random variable $N=\epsilon_1 W(v_1)+\ldots+\epsilon_k W(v_k)$. Write this as $N_1+N_2$, where $N_1=\sum_{i\le i_0}\epsilon_i W(v_i)$ and $N_2=\sum_{i>i_0}\epsilon_i W(v_i)$; and $i_0$ is chosen so that $W(v_i)=1$ if and only if $i\le i_0$.

Now $\mathbb P(N_1+N_2=\ell)\le \sup_r \mathbb P(N_1+N_2=\ell|N_2=r)
\le \mathbb P(N_1=\lfloor i_0/2\rfloor)=2^{-i_0}\binom{i_0}{\lfloor i_0/2\rfloor}$.
This is known to be approximately $\sqrt {2/\pi i_0}\le 1.01 \sqrt{2/\pi n}$.