For a given element do heavier isotopes form stronger bonds than lighter isotopes?
Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by; 
\begin{equation}
\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}
\end{equation}
Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega $ with  $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2. 

So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well. 

In vibrational spectroscopy this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.

Edit

I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the uncertainty principle in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)

The quantum oscillator has a Hamiltonian of the classical oscillator except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.

The classical Hamiltonian (or total energy) is learnt at school to be kinetic + potential,
\begin{equation}
H=\frac{p^2}{2\mu}+\frac 12 kx^2
\end{equation} 
For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by $\hat p$ and $\hat x$ where,
\begin{equation}
\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x
\end{equation}
Therefore with these changes in the above classical expression we generate the quantum Hamiltonian of the quantum harmonic oscillator, 
\begin{equation}
\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2 
\end{equation}
Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.

It is another principle of quantum mechanics that the Schrodinger equation is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form. 
\begin{equation}
\hat H\psi =E\psi
\end{equation}
Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get, 

\begin{equation}
E_n=(n+\frac 12)\hbar \omega
\end{equation}
What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution. 

Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$). 

This is a zero point energy. 

As promised here is the home-made graph of the molecular potential energy. Im sorry its not very good but honestly the best I could do with the available tools! Apologies. 

The main points to follow are:


At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number). 
Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case. 
To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines.