Such a bound can be derived with Girsanov's theorem and Pinsker's inequality. Let $X_t = x_0 + W_t$. Supposing $b$ has linear growth in $x$, we may define measures $P_i$, $i=1,2$ by

$\frac{dP_i}{dP} = \exp\left(\int_0^Tb_i(t,X_t)dW_t - \frac{1}{2}\int_0^T|b_i(t,X_t)|^2dt\right)$.

Then, under $P_i$, $W^i_t := W_t - \int_0^tb_i(s,X_s)ds$ is a Brownian motion, and so $X$ is a weak solution of SDE (i). Let $P^i \circ X^{-1}$ denote the $P^i$-law of the entire process, a measure on the space of continuous functions. Then

$\frac{dP_2}{dP_1} = \exp\left(\int_0^T(b_2(t,X_t) - b_1(t,X_t))dW^1_t - \frac{1}{2}\int_0^T|b_2(t,X_t) - b_1(t,X_t)|^2dt\right)$.

Let $\mathcal{H}(\cdot | \cdot)$ denote the relative entropy. By Pinsker's inequality,

$
\begin{align}
d_{TV}^2(P_1 \circ X^{-1}, P_2 \circ X^{-1}) &= d_{TV}^2(P_1, P_2) \le 2\mathcal{H}(P_1 | P_2) \newline
&= -2\mathbb{E}^{P_1}\left[\log \frac{dP^2}{dP^1}\right] \newline
&= \mathbb{E}^{P_1}\left[\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right] \newline
&= \mathbb{E}^{P}\left[\frac{dP_1}{dP}\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right]
\end{align}
$

since the stochastic integral term is a true martingale. This is actually a much stronger control than you requested, since it is easy to see that $d_{TV}(P_1 \circ X_t^{-1}, P_2 \circ X_t^{-1}) \le d_{TV}(P_1 \circ X^{-1}, P_2 \circ X^{-1})$ for any $t$. If you have a uniform bound on $|b_1 - b_2|$, you have a good bound on the $TV$ distance between your processes. You can probably get away with an $L^2$ bound, but you'll then need to fuss with the $dP_1/dP$ term a bit. Hope this helps!

EDIT 1: It's interesting to note that this approach breaks down if you have two different volatility coefficients in your SDE, because the laws of the processes are then singular. But you could still bound the TV distance between the time-$t$ laws, as requested, probably via Malliavin calculus.

EDIT 2: I should elaborate on the first of the string of equalities above. Since $X$ and $W$ generate the same $\sigma$-fields, $dP^i/dP$ is $X$-measurable, and so

$\frac{dP^i \circ X^{-1}}{dP \circ X^{-1}} (X) = \mathbb{E}\left[\frac{dP^i}{dP}|X\right] = \frac{dP^i}{dP}$.

From this it is clear that the TV-distances above are the same, from the formula $d_{TV}(\mu,\nu) = \int d\lambda|d\mu/d\lambda - d\nu/d\lambda|$ for $\mu,\nu \ll \lambda$.