My first answer was wrong. The answer is no.

First observe that the embedding $X\to \mathcal D\times\mathbb P^2$ is a bit distracting. The map $X\to \mathbb P^2$ plays no role in the question of transversality to a given $\tilde H_q$. So I think about the problem like this:

You have smooth maps

$\mathbb P^2\leftarrow H\to \mathcal D\leftarrow X$,

where $H\subset \mathcal D\times \mathbb P^2$ is as in my wrong answer (i.e. the space of all pairs $(f,q)$ where $q$ is on the curve defined by $f$). The projection $H\to \mathbb P^2$ is a submersion, so the fiber $H_q$ over any $q$ is a manifold, and the question is whether for a dense set of $q$ the map $H_q\to \mathcal D$ is transverse to the map $X\to \mathcal D$.

(We say that two maps $A\to B\leftarrow C$ are transverse if whenever points $a$ and $c$ both go to the point $b$ then the tangent space $T_bB$ is spanned by the images of $T_aA$ and $T_cC$.)

If the map $H\to \mathcal D$ were a submersion, then the answer would be yes (for any smooth $X$ and any map $X\to \mathcal D$), by the following argument:

The submersion $H\to \mathcal D$ is transverse to any $X\to \mathcal D$. Thus the fiber product $Y=H\times_{\mathcal D}X$ is a manifold, and a little bit of playing with tangent spaces yields that those $q$ for which $H_q\to \mathcal D$ is transverse to $X\to \mathcal D$ are precisely the regular values of the projection $Y\to \mathbb P^2$. In particular the transversality holds for a dense set of choices of $q$.

But $H\to \mathcal D$ is not a submersion; this fails at precisely those points $(f,q)$ such that $q$ is a non-smooth point of the curve defined by $f$.

And there are counterexamples with $d=2$. Let's work in an affine plane in $\mathbb P^2$ with coordinates $(x,y)$. For each $q=(x_0,y_0)$ the quadratic equation $(x-x_0)(y-y_0)=0$ defines a curve through $q$. Let $X\subset \mathcal D$ be this two-dimensional family. For any $(x_0,y_0)$ the manifold $H_q$ intersects $X$ non-transversely.