It depends what you mean by "compatible." For any Z-form of a finite-dimensional C-algebra, there's a canonical Z-form for any quotient just given by the image (the image is a finitely generated abelian subgroup, and thus a lattice). I'll note that the integral form Bruce suggests below is precisely the one induced this way by the Kazhdan-Lusztig basis, since his presentation is the presentation of the Hecke algebra via the K-L basis vectors for reflections, with the additional relations.
What you could lose when you take quotients is positivity (which I presume is one of things you are after). The Hecke algebra of S_n has a basis so nice I would call it "canonical" but usually called Kazhdan-Lusztig. This basis has a a very strong positivity property (its structure coefficients are Laurent polynomials with positive integer coefficients). I would argue that this is the structure you are interested in preserving in the quotient.
If you want a basis of an algebra to descend a quotient, you'd better hope that the intersection of the basis with the kernel is a basis of the kernel (so that the image of the basis is a basis and a bunch of 0's). An ideal in the Hecke algebra which has a basis given by a subset of the KL basis is called "cellular."
The kernel of the map to TLd, and more generally to EndU_q(sl_n)(V⊗d) for any n and d, is cellular. Basically, this is because the parititions corresponding to killed representations form an upper order ideal in the dominance poset of partitions.
However, the kernel of the map to STLd is not cellular. In particular, every cellular ideal contains the alternating representation, so any quotient where the alternating representation survives is not cellular. So, while STLd inherits a perfectly good Z-form, it doesn't inherit any particular basis from the Hecke algebra.
I'm genuinely unsure if this is really a problem from your standpoint. I mean, the representation V⊗d still has a basis on which the image of any positive integral linear combination of KL basis vectors acts with positive integral coefficients. However, I don't think this guarantees any kind of positivity of structure coefficients. Also, Stroppel and Mazorchuk have a categorification of the Artin-Wedderburn basis of S_n, so maybe it's not as bad as you thought.
Anyways, if people want to have a real discussion about this, I suggest we retire to the nLab. I've started a relevant page there.
The name of the concept you are looking for is the Schur index. The Schur index is $1$ iff the representation can be realized over the field of values. The Schur index divides the degree of the character.
In your case, the the Schur index is either $1$ or $2$. You can use a variety of tests to eliminate $2$, but for instance:
Fein, Burton; Yamada, Toshihiko, The Schur index and the order and exponent of a finite group, J. Algebra 28, 496-498 (1974). ZBL0243.20008.
shows that if the Schur index was $2$, then $4$ divides the exponent of $G$.
In other words, all of your representations are realizable over the field of values.
Isaacs's Character Theory of Finite Groups has most of this in it, and I found the rest of what I needed in Berkovich's Character Theory collections. Let me know if you want more specific textbook references.
Edit: I went ahead and looked up the Isaacs pages, and looks like textbook is enough here: Lemma 10.8 on page 165 handles induced irreducible characters from complemented subgroups, and shows that the Schur index divides the order of the original character. Taking the subgroup to be the rotation subgroup and the original character to be faithful (or whichever one you need for your particular irreducible when $n$ isn't prime), you get that the Schur index divides $1$. The basics of the Schur index are collected in Corollary 10.2 on page 161.
At any rate, Schur indices are nice to know about, and if Isaacs's book doesn't have what you want, then Berkovich (or Huppert) has just a silly number of results helping to calculate it.
Edit: Explicit matrices can be found too. If $n=4k+2$ is not divisible $4$, and $G$ is a dihedral group of order $n$ with presentation $\langle a,b \mid aa=b^n=1, ba=ab^{n-1} \rangle$, then one can use companion polynomials to give an explicit representation (basically creating an induced representation from a complemented subgroup). Send $a$ to $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, also known as multiplication by $x$. Send $b$ to $\begin{pmatrix}0 & -1\\1 & \zeta + \frac{1}{\zeta}\end{pmatrix}$, also known as the companion matrix to the minimum polynomial of $\zeta$ over the field $\mathbb{Q}(\zeta+\frac{1}{\zeta})$, where $\zeta$ is a primitive $(2k+1)$st root of unity.
Compare this to the more direct choice of $a = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ and $b = \begin{pmatrix} \zeta & 0\\0 & \frac{1}{\zeta}\end{pmatrix}$. If you conjugate this by $\begin{pmatrix}1 & \zeta \\\zeta & 1\end{pmatrix}$ then you get my suggested choice of a representation.
In general, finding pretty, (nearly-)integral representations over a minimal splitting field is hard (and there may not be a unique minimal splitting field), but in some cases you can do it nicely.
Let me know if you continue to find this stuff interesting. I could ramble on quite a bit longer, but I think MO prefers focused answers.
Best Answer
In the following paper the authors deal with the Wedderburn decomposition of group algebras of finite metacyclic groups over a finite field:
G.K. Bakshi - S. Gupta - I.B. Passi: Semisimple metacyclic group algebras, Proc. Indian Acad. Sci., Math. Sci. 121, No. 4, 379-396 (2011).
It is available at this link:
http://www.ias.ac.in/mathsci/vol121/nov2011/pmsc-d-10-00210.pdf
The Wedderburn decomposition you are looking for is discussed as an example in Section 5 of the mentioned paper.